Antibonding MOs place more electron density near the less electronegative atom in the diatomic molecule. If we look at the valence shell of Xe there are a total of six electrons in the 5p orbital and two electrons in the 5s orbital. Provide the following information for the compound NH3. The shape of a molecule can be determined using valence shell electron pair repulsion (VSEPR) theory. What is the hybridization of the sulfur atom, electronic geometry, and molecular geometry in SF4 molecule? Therefore, then mark those two electrons pairs on xenon atom. The number of pd 'pi' bonds present in XeO 3 and XeO 4 molecules, respectively are A 3,4 B 4,2 C 2,3 D 3,2 Medium Solution Verified by Toppr Correct option is A) Structure of XeO 3 has 3pd bonds. (ICl2)1- 6.) This is to determine the number of electrons further required to complete the octet of the connected atoms. More answers below Parul Goyal M.Sc. Therefore, XeF4 molecular geometry is square planar. Therefore, XeOF4 is a polar molecule. In the case of fluorine, four F atoms bond with these four half filled orbitals. CH3COCH3 Lewis Structure, Molecular Geometry, Hybridization. It occurs only when the atoms undergo the process of bond formation and not when they are in their gaseous state. I need to determine: 1.) Also, the inter-electronic repulsion influence the bond angles between different atoms in a molecule. Im a mother of two crazy kids and a science lover with a passion for sharing the wonders of our universe. {/eq}, identify the following: name, number of valence electrons, number of electron domains, parent geometry, molecular geometry, hybridization, number of sigma bonds, and number of pi bonds. In terms of solubility, it is completely miscible in water and ethanol. Similar to other oxides of Xenon it is also very unstable and highly reactive. As per this rule, an atom becomes most stable when it has eight electrons in its valence shell. All rights reserved.
sp2 Hybridization | Introduction to Chemistry - Course Hero Continue with Recommended Cookies. The formal charge is an important indicator that confirms the existence and identifies the credibility of lewis structure. This results in sp3d2 hybridization. However, we also notice that all the atoms of this molecule contain one or more lone pairs of electrons. Provide the following information for the compound CH2Cl2. This is also known as the rule of eight. To achieve octet stability 4 fluorine atoms will share their 1 electron each with a Xenon atom thereby providing the desired XeF4 lewis structure. valence electrons given by fluorine atoms =, There are already 4 sigma bonds in the above drawn basic sketch. Both the p y and the p z orbitals on each carbon atom form pi bonds between each other. Indicate the number of sigma bonds and the number of pi bonds in the molecule CO2. A pi bond between two atoms is formed only in addition to a sigma bond. =CH?C? The electronegativity difference is 1.4. How many lone pairs are there in SF 6 lewis structures? c) Determine the total number of sigma and pi bonds in the molecule. Step 3: Next, all the atoms are joined with the central atom using a single bond.
XeF4 Lewis Structure: Drawings, Hybridization, Shape, Charges, Pair a. The molecular geometry and shape are very important in analyzing reactivity, polarity, color, and the other associated properties. This is the reason for the square planar structure of the compound xenon tetrafluoride. The carbon-carbon triple bond is the shortest and is sp hybridised. Continue Reading Lawrence C. FinTech Enthusiast, Expert Investor, Finance at Masterworks Wed (See the HF molecule in Figs. xenon atom. By accepting all cookies, you agree to our use of cookies to deliver and maintain our services and site, improve the quality of Reddit, personalize Reddit content and advertising, and measure the effectiveness of advertising. The molecule XeF4 is nonpolarIII. a) What is hybridization of valence shell orbitals are predicted on sulphur or carbon in SO2 and CO2?
Which of the following statements is (are) incorrect? I) The The nickel used in this process does not act as a catalyst. The number of electron domain will dictate the hybridization and the parent and molecular geometry using VSEPR theory.
IB Chemistry on Valence Bond and Hybridization Theory Lewis Structure Ch2chch3 - qhaw.ayuntamientobalboa.es Number of valence electrons for molecule {Blank} 2. Now we know how many electrons includes in valence shells of xenon and fluorine atom. For the compound {eq}XeF_4 G. N. Lewis is well known for multiple contributions to the study of chemistry. The molecules are: 1.) Number of bonds: 11 single bonds II sigma bonds,..H.0. What is the lewis structure for GeCl_6-2, determine the hybridization on the center atom and what are the sigma and pi bonds? The stable XeF4 lewis structure can only be obtained if the lone pair on xenon are in perpendicular alignment in an octahedral geometry. See Answer.
How to Find Hybridization | Shape | Molecule | Adichemistry Some of the important properties are discussed below in detail. What is the hybridization of the sulfur atom, electronic geometry, and molecular geometry in SF5^+ ion? Check the stability and minimize charges on atoms by converting lone pairs to bonds until most stable Identify the type of bonds created by the orbitals involved in the bonds labeled c and d. For the following molecule, identify the central atom, the steric number on the central atom, the number of bonded electron pairs on the central atom, and the number of lone pairs on the central atom: Nitrogen trifluoride. (Image), Predict the electron pair geometry and the molecular structure for the polyatomic ion: BrCl_4^-. The number of sigma bond is 4 and the number of lone pairs is 0. As both xenon and fluorine in the XeF4 lewis structure are non-metals so there will be sharing of electrons. 3 electron groups = sp2, 4 groups = sp3, 5 groups = sp3d, 6 groups = sp3d2. Figure 9.24. Two valence electrons remain on the Xenon atom as a lone pair. a. How many sigma and pi bonds are in the Lewis structure for ch2chch3? The triple bond contains one sigma bond and two pi bonds. (a) are confined between two adjacent bonding atoms (b) are free to move around the six-membered ring. What are the conditions necessary for oxidation?Answer my questions in shortI can mark you as BRAINLIEST , Explain the preparation of HSO4 by contact process., Calculate the molecular weight of Ca(NO), answer the following questions: Why do atoms combine to form molecules?. In the structure, it is clear that four valence electrons form single bonds with four fluorine atoms. In order to exhibit delocalized pi bonding, a molecule must have ____.
(DOC) 1973 D - Science Done Wrightsciencedonewright.weebly.com/uploads If you need more explanation or help say so. Sigma bonds are formed by end-to-end overlapping and Pi bonds are when the lobe of one atomic orbital overlaps another. The symmetry of the molecules also influences their polarity. For CF3Cl: How many bonding pairs? E) sp^3, trigonal planar, 120 degrees. A pi bond (VB or MO theory) or a double bond (VSEPR theory) does not affect the geometry of a molecule. Sigma and pi bonds are formed by the overlap of atomic orbitals. To complete its octet stability criteria each fluorine atom will share its 1 electron with the central xenon atom. However, Xenon is seen to have more than eight valence electrons i.e. In the XeF4 lewis structure, there is the presence of fluorine atoms which belong to the halogen family and have 7 valence electrons. Now only, Usually as a theory, those remaining electron pairs should be first marked on outside atoms. To clarify, a double bond consists of one sigma bond and one pi bond and a central atom with one double bond will have sp2 hybridization. The energy is redistributed to the other orbitals to give equivalent energy to those orbitals. Step 7: Calculating the formal charge on XeOF4 molecule: For Xenon atom, Total number of valence electrons in free state = 8, Total number of non-bonding electrons = 2, Therefore, Formal charge on Xenon atom = 8 2 (12), For Oxygen atom, Total number of valence electrons in free state = 6, Therefore, Formal charge on nitrogen atom = 6 4 (4), For Fluorine atom, Total number of valence electrons in free state = 7, Therefore, Formal charge on nitrogen atom = 7 6 (2). Bond angles = _ What is the hybridization of the central atom in POC, In the molecule given below, find the number of (a) pi bonds (b) sigma bonds. Talking about the XeF4 lewis structure then it does not show the resonance phenomenon because it is an asymmetrical structure and does not have double bonds for the movement. Explanation: The structure of is given in the image below. The same level atomic orbitals are allowed to take part in the process. Also, we know that Xenon in itself has sp3 hybridization state and contains 8 valence electrons. There are three lone pairs on each fluorine atom. D-3.1, 1.25. Save my name, email, and website in this browser for the next time I comment. Bond Angle of SH 2 (Answer): Less than 109. The structure entails the goal of obtaining an electronic configuration, which also means that it follows the octet rule and fits appropriately to the molecules to form the structure. Ideal Bond Angles : Hybridi zation : Example : 2 : linear : 0 "linear" 180 sp : HCN. If you would like to change your settings or withdraw consent at any time, the link to do so is in our privacy policy accessible from our home page.. Hybridization is a process where orbitals of the atoms involved in molecule formation intermix and form new hybrid orbitals with distinguished properties. a. Lewis structure b. number of valence electrons c. number of electron domains (both bonding and nonbonding) d. electron domain geometry e. molecular geometry. The Lewis Dot structure starts with three C atoms singly bonded in a . Also, there is the presence of a single covalent bond between Xe and F which is also called the sigma bond. a) II is incorrect.
There shall be six electrons around each fluorine atom. Join / Login >> Class 12 >> Chemistry >> The p-Block Elements >> Compounds of Xenon >> Number of bonds, pi bonds and lone pai. What would be the molecular geometry of this molecule? For the compound XeF 4 X e F 4, identify the following: name, number of valence electrons, number of electron domains, parent geometry, molecular geometry, hybridization, number of sigma bonds,. IF_5, Provide the following information for the compound H2S. So xenon will be the central atom surrounded by 4 fluorine atoms. Choose from the following word bank: tetrahedral, trigonal pyramidal T-shaped, trigonal planar octahedral, trigonal planar trigonal, What is the hybridization of the central atom in BCl_3? These fluorine atoms will then be placed on either side of the central atom. Molecular Geometry This problem has been solved! Here, all the angles should be less than 90 due to the repulsion forces that exist between the lone pairs as well as bond pairs of different atoms. Choice (B) counts the double bonds as 2 pi bonds rather than I sigma and I pi bond. Bond angles of 90 degrees B. Resonance structures C. Expanded octet of electrons D. dsp3 hybridization of orbitals E. Elect, Give the following information for H3O+. State the number of electron groups and lone pairs and use VSEPR theory to determine the shape of the following molecules or ions as tetrahedron, trigonal pyramidal , or bent. How many valence electrons occupy sigma bond orbitals an pi bond orbitals in c14h10? Now if we follow the VSEPR theory, the net electronic repulsions has to be minimum. Get access to this video and our entire Q&A library. To find hybridization in general think of it like this: A hybrid orbital needs to form whenever there's a bond or a lone pair, we call either of these an electron group. a. PF3 b. H2O c. CCl4. Name the structure of (Co (H_2 O)_3 Cl (NH_3)_2)^{2+} 1) tetrahedral 2) square planar 3) pentagonal bipyramidal 4) octahedral 5)linear, Identify the hybridization of each carbon atom in the following molecule. How many electron groups are on the central atom? CS_2 \\d. a. tetrahedral b. trigonal planar c. trigonal pyramidal d. trigonal bipyramidal e. seesaw. What is the Lewis formula? To make this compound, one part of xenon is mixed with exactly four parts of the element, fluorine. How many sigma and pi bonds are present in H2CO? Linear C. Trigonal bipyramidal D. Trigonal planar E. See-saw, Identify the hybridization of each underlined atom in the following compound: H2C?? It vigorously reacts with water to form extremely corrosive and dangerous compounds that tend to explode and hence, it should always be stored away from water. Xe-F bonds are all sigma bonded. In simple and mathematical words if the formal charge of each atom in the molecule is 0 then the molecule is identified according to the criteria of the lewis structure. Search: Ch3coch3 Lewis Structure. After finding the number of valence electrons participating in the XeF4 lewis structure, it is important to look out for the central atom. Step 6: The final step towards authentication of a derived Lewis structure is the calculation of formal charge. D) sp^2, linear, 120 degrees. //]]>. Hence XeF4 lewis structure has sp3d2 hybridization. First, the Lewis structure of the compound is drawn. It has a distance of 1.95 A between the Xe and F. This is particularly explained by the valence bond theory, as the two non-bonding electrons from 5p are promoted to the 5d orbital. A molecule with 3 single bonds and 0 lone pair of electrons is predicted to have which type of molecular geometry? b) I and IV are. The remaining two excited electrons bond with the oxygen atom forming a sigma and a pi bond, respectively. lewis structure. An sp-hybridized atom possesses s orbital(s), p orbital(s), and hybrid orbital(s) in its valence shell. a. trigonal planar b. trigonal pyramidal c. bent d. tetrahedral, Describe the hybrid orbitals used by the underlined atoms in the molecule below and then provide the number of bonds (sigma and pi) used by each of the atoms. A Lewis structure represents the location of valence electrons around the atoms of a molecule in a pictorial form. See Answer Show transcribed image text Expert Answer 100% (2 ratings)
XeF4 Lewis Structure, Molecular Geometry - Techiescientist XeF4: 4 single bonds, 2 lone pairs (another expanded octet, in this case the formal charge on the central atom can reach 0 by putting 2 lone pairs on it, Xenon . The sublimation of the compound occurs at 115.7C. What is the hybridization for each C atom and for each O or N atom? The molecule XeF4 is nonpolarIII. What is the hybridization of the central atom in XeF4? 100% (4 ratings) XeF4 shape is square planner. Is it polar or non polar? So XeF4 all properties are explained by a single structure that has a square planar shape. It helps to figure out the number of bonding and a non-bonding group of electrons. The electrons of an atom that revolve farthest from its nucleus in its outermost orbit are known as valence electrons. BrF5. What is VSEPR theory? Because there are no double or triple bond, there is no pi bond in the compound. A. octahedral B. tetrahedral C. trigonal planar D. trigonal pyramidal E. trigonal bipyramidal, Provide the following information for the compound BrF5. The bond angles are less than 90. The molecule XeF4 is a nonpolar molecule. how many individual sigma bonds and individual pi bonds are in XeF4 sigma bonds: pi bonds: Expert Answer Who are the experts? is a group IA element and has 8 electrons in its last shell (valence shell). (Hint: Draw the lewis dot structure of each To use the VSEPR model to determine the geometry of a molecule we need to know which of the following? In this tutorial, we will learn how to draw lewis structure What is the hybridization of the sulfur atom, electronic geometry, and molecular geometry in SO3^2 ion? Therefore, 7*4 shall give us 28; the total valence electrons of XeF4 come to be 8+28 which is 36. Now calculating the total number of electrons in XeOF4 molecule: Step 2: Now, a central atom is chosen. It also helps to identify the strength between the bonds and the electronic transitions.
13.2. Molecular orbitals for ethene | Organic Chemistry II These orbitals are one s, three p, and two d orbitals. Bond angles(<,>,or ideal) 2.) Two orbitals of an atom having the same energy level fuse together to form a new kind of an orbital. It is important to understand the concept of valence electrons, as these participate in the formation of bonds between different atoms by acquiring, losing, or sharing electrons. a. Lewis dot structure b. hybridization c. electron geometry d. molecular geometry e. polarity. oxygen and fluorine, are more electronegative than it. Create the Lewis structure. A molecule with the formula AB_3 has a trigonal planar geometry. lone pairs on atoms. Create an account to follow your favorite communities and start taking part in conversations. has 8 valence electrons. Xenon atom is the center atom and each fluorine atom has made a single bond with xenon atom. VSEPR theory predicts the distribution of the electron domains around a central atom by arranging them in a way that the repulsion between these electrons is minimized. However, the geometry within that of the molecules is not explained by the Lewis structure method. In the XeF4 lewis structure, two orbitals namely fluorine and xenon are involved. Coming back to the XeF4 lewis structure then there are 4 lone pairs or nonbonding electrons present on the central xenon atom. Between Xenon and fluorine, xenon is the least electronegative element and fluorine is the one with the highest electronegativity. (a) The -bonded framework is formed by the overlap of two sets of singly occupied carbon sp2 hybrid orbitals and four singly occupied hydrogen 1s orbitals to form electron-pair bonds. a) trigonal planar b) trigonal bipyramidal c) octahedral d) tetrahedral e) trigonal pyramidal. 14. Dots are used to characterize the electrons and lines are drawn to show the bonds between two atoms. C) sp, trigonal planar, 120 degrees. Now, it can be seen that there are single bonds that are four in number. Therefore, xenon becomes the center atom and each fluorine atom is joint with xenon atom. As per this theory, the numbers of bond pairs, as well as lone pairs of electrons, are majorly involved in devising the shape of a molecule. For the compound XeF2, identify the following: name, number of valence electrons, number of electron domains, parent geometry, molecular geometry, hybridization, number of sigma bonds, and number of pi bonds. It mostly depends upon the electronegativity of atoms as the more electronegative species tends to pull the shared pair of electrons towards itself, hence, creating an imbalance in charge distribution. (a) SF_6 (b) IF_5 (c) XeF_4, If a compound has two nonbonded pairs of electrons in its Lewis structure, what is its molecular geometry? XeF. Step 8: Therefore, the net formal charge on the XeOF4 molecule is zero indicating that the above derived Lewis structure is most accurate for this molecule. We review their content and use your feedback to keep the quality high. (Triple bond = 1 sigma + 2 pi bonds). As such XeF4 lewis structure does not have major applications. This is a rare example of a noble gas forming a chemical compound. Quantum mechanics prove that the hybridized orbitals are the deciding factor in the geometry of the molecule. This is generally helpful in showcasing the arrangement of electrons that lie around an individual atom in the specific molecule. In the case of the XeOF4 molecule, Xenon is a noble gas and a group 18th element has 8 valence electrons, Oxygen is a group 16 element and had 6 electrons in its outermost shell while fluorine is group 17 element having 7 valence electrons. XeF4 is the chemical formula of the compound Xenon Tetrafluoride. If you are a beginner to lewis structure drawing, follow these sections slowly and properly to understand Now if we follow the VSEPR theory, the net electronic repulsions has to be minimum.
Full | PDF However, the oxygen atom still requires one more electron to complete its octet. The bond angle between the 2 lone pairs of electrons is 180. structure is obtained. is given below. What is are the bond angles at the central atom of the molecule. The molecular geometry of a compound is determined by using the Valence Shell Electron Pair Repulsion Theory. A double bond contains a sigma bond and a pi bond. a. ?H3, Predict the electron-pair geometry and the molecular geometry of sulfur dioxide, XeF_4. A crystalline substance that is colorless in nature, is formed when xenon reacts with fluorine. The number of sigma and pi-bonds are eight and one respectively. Therefore, there are twelve electrons in the valence shell of sulphur in SF 6. it can house more than eight electrons in its valence shell.
pw Module 3 | PDF | Periodic Table | Electron Configuration The purpose of this subreddit is to help you learn (not complete your last-minute homework), and our rules are designed to reinforce this. With two lone pairs, the 4F atoms will be positioned in one plane and thus, the molecular geometry is square planar. AIH3 2.) It's late, but I did it. Sigma bonds are single bonds. How many sigma and pi bonds are present in the following molecule? a. octahedral b. trigonal bipyramidal c. square pyramidal d. tetrahedral, According to the VSEPR model, a molecule with the general formula AB5 with no lone pairs on the central atom will have a __________ molecular geometry. Xenon.
XeOF4 Lewis Structure, Geometry, Hybridization, and Polarity Total electron pairs are determined by dividing the number total valence electrons by two.
XeF4 (Xenon tetrafluoride) Lewis Structure - chemistryscl.com [CDATA[
Solved how many individual sigma bonds and individual pi | Chegg.com Usually. Fluorine is a group VIIA element in the periodic table and contains 7 electrons in their last shell. The Lewis structure of the compound is shown below: There are 6 electron domains around the central atom F (4 single bonds and 2 lone pairs). It can be prepared by reacting XeF6 with NaNO3 or silica. (a) 3 bonding pairs and 1 lone pair (b) 2 bonding pairs and 3 lone pairs (c) 2 bonding pairs and 2 lone pairs (d) 4 bonding pairs and 0 lone pairs (e) 1 bonding pair and 3 lone. Step 1 We need to count the valence electrons of the xenon tetrafluoride molecule with the help of a periodic table.
Sigma and Pi bonds - CHEMISTRY COMMUNITY - University of California The lone pair of electrons generates electronic repulsion which needs to be minimized for stability according to the VSEPR theory. Number of bonded electron pairs on central atom {Blank} 3. cannot be a center atom because fluorine's maximum valence is 1 (fluorine cannot make two or more bonds). History and many academic research papers confirm that the first compound of xenon was not XeF4 lewis structure but Xe+ PtF6- as reported by scientist Bartlett. Your email address will not be published. For, It is assumed that all the other atoms of a molecule are connected with the central atom. Comedian Joe Lycett shreds 10,000 in protest over David Beckham's Qatar deal Comedian Joe Lycett has shredded 10,000 of his own money after he presented David Beckham with an ultimatum to pull out of his role as a Qatar ambassador or he would destroy the cash. Polar or nonpolar. a. total number of valance electrons b. name of the electron geometry c. number of bonds attached to the central atom d. number of electron pairs on the central atom e. name of the molecular geometry. You'll get a detailed solution from a subject matter expert that helps you learn core concepts. Anyways, im studying for my chem final right now, so I might finish off the rest of them.
clf polar or nonpolar atom closest to negative side Use the MO model to predict which atom of carbon monoxide should form bonds to metals. in Chemistry & Organic Chemistry, Jamia Hamdard (Graduated 2017) Author has 219 answers and 497.7K answer views 4 y Once the form for bonds, we need to put three bonds in between here now. What are the approximate bond angles in this substance? a. total number of valance electrons b. name of the electron geometry c. number of bonds attached to the central atom d. number of electron pairs on the central atom e. name of the molecular geometry, Fill in the blanks. SF4 has . Total valance electrons
Preparing For Your Acs Examination In General Chemistry Second Edition So this confirms the uniform electric charge distribution and the existence of the XeF4 lewis structure. The sp hybrid orbitals form a sigma bond between each other as well as sigma bonds to the hydrogen atoms. Ethene Answer: XeF4 11 In PO34 ion, the formal charge on each oxygen atom and P-O bond order respectively are A-0.75, 0.6.
[College Chemistry] Molecular Geometry : HomeworkHelp - reddit Xenon Oxytetrafluoride is a colorless inorganic compound. Hence, forming XeF4 molecule, in which there are two lone pairs and four bond pairs. However, when it comes in contact with water, it reacts with it and releases xenon gas, molecular oxygen, and xenon fluoride. In order to achieve this, the lone pairs lie in a perpendicular plane in an octahedral arrangement opposite (180 degree) from each other.
How to calculate the number of sigma and pi bonds - Quora window.__mirage2 = {petok:"zRB_SPG4DyiJSUD8fc4gefmbgQuc.KSr9hqBRRTgqgA-31536000-0"}; The valence shell of the atom contains 6 electrons in the 5p orbital whereas the 5s orbital entails 2 electrons. But the XeF4 lewis structure is square planar which is symmetrical. XeF4 lewis structure exhibits the property of sublimation at room temperature.
How many sigma and pi bond in xef4? - Answers oxygen and fluorine, are more electronegative than it. NCERT Solutions Class 12 Business Studies, NCERT Solutions Class 12 Accountancy Part 1, NCERT Solutions Class 12 Accountancy Part 2, NCERT Solutions Class 11 Business Studies, NCERT Solutions for Class 10 Social Science, NCERT Solutions for Class 10 Maths Chapter 1, NCERT Solutions for Class 10 Maths Chapter 2, NCERT Solutions for Class 10 Maths Chapter 3, NCERT Solutions for Class 10 Maths Chapter 4, NCERT Solutions for Class 10 Maths Chapter 5, NCERT Solutions for Class 10 Maths Chapter 6, NCERT Solutions for Class 10 Maths Chapter 7, NCERT Solutions for Class 10 Maths Chapter 8, NCERT Solutions for Class 10 Maths Chapter 9, NCERT Solutions for Class 10 Maths Chapter 10, NCERT Solutions for Class 10 Maths Chapter 11, NCERT Solutions for Class 10 Maths Chapter 12, NCERT Solutions for Class 10 Maths Chapter 13, NCERT Solutions for Class 10 Maths Chapter 14, NCERT Solutions for Class 10 Maths Chapter 15, NCERT Solutions for Class 10 Science Chapter 1, NCERT Solutions for Class 10 Science Chapter 2, NCERT Solutions for Class 10 Science Chapter 3, NCERT Solutions for Class 10 Science Chapter 4, NCERT Solutions for Class 10 Science Chapter 5, NCERT Solutions for Class 10 Science Chapter 6, NCERT Solutions for Class 10 Science Chapter 7, NCERT Solutions for Class 10 Science Chapter 8, NCERT Solutions for Class 10 Science Chapter 9, NCERT Solutions for Class 10 Science Chapter 10, NCERT Solutions for Class 10 Science Chapter 11, NCERT Solutions for Class 10 Science Chapter 12, NCERT Solutions for Class 10 Science Chapter 13, NCERT Solutions for Class 10 Science Chapter 14, NCERT Solutions for Class 10 Science Chapter 15, NCERT Solutions for Class 10 Science Chapter 16, NCERT Solutions For Class 9 Social Science, NCERT Solutions For Class 9 Maths Chapter 1, NCERT Solutions For Class 9 Maths Chapter 2, NCERT Solutions For Class 9 Maths Chapter 3, NCERT Solutions For Class 9 Maths Chapter 4, NCERT Solutions For Class 9 Maths Chapter 5, NCERT Solutions For Class 9 Maths Chapter 6, NCERT Solutions For Class 9 Maths Chapter 7, NCERT Solutions For Class 9 Maths Chapter 8, NCERT Solutions For Class 9 Maths Chapter 9, NCERT Solutions For Class 9 Maths Chapter 10, NCERT Solutions For Class 9 Maths Chapter 11, NCERT Solutions For Class 9 Maths Chapter 12, NCERT Solutions For Class 9 Maths Chapter 13, NCERT Solutions For Class 9 Maths Chapter 14, NCERT Solutions For Class 9 Maths Chapter 15, NCERT Solutions for Class 9 Science Chapter 1, NCERT Solutions for Class 9 Science Chapter 2, NCERT Solutions for Class 9 Science Chapter 3, NCERT Solutions for Class 9 Science Chapter 4, NCERT Solutions for Class 9 Science Chapter 5, NCERT Solutions for Class 9 Science Chapter 6, NCERT Solutions for Class 9 Science Chapter 7, NCERT Solutions for Class 9 Science Chapter 8, NCERT Solutions for Class 9 Science Chapter 9, NCERT Solutions for Class 9 Science Chapter 10, NCERT Solutions for Class 9 Science Chapter 11, NCERT Solutions for Class 9 Science Chapter 12, NCERT Solutions for Class 9 Science Chapter 13, NCERT Solutions for Class 9 Science Chapter 14, NCERT Solutions for Class 9 Science Chapter 15, NCERT Solutions for Class 8 Social Science, NCERT Solutions for Class 7 Social Science, NCERT Solutions For Class 6 Social Science, CBSE Previous Year Question Papers Class 10, CBSE Previous Year Question Papers Class 12, JEE Advanced Previous Year Question Papers, JEE Main Chapter-wise Questions and Solutions, JEE Advanced Chapter-wise Questions and Solutions, JEE Main 2023 Question Papers with Answers, JEE Main 2022 Question Papers with Answers, JEE Advanced 2022 Question Paper with Answers.