What is the approximate value of the bond angle indicated? Identify the nonpolar

Question

What is the approximate value of the bond angle indicated? Identify the nonpolar molecules in the following group: SO_2, NH_3, Which of the following atoms, in its ground state, does not have unpaired electrons? (Sketch all that apply.) According to valence bond theory, how many bonds would you expect a nitrogen atom (in its ground state) to form? How many orbitals docs a set of sp2 hybrid orbitals contain? How many p atomic orbitals arc required to generate a set of sp^3 hybrid orbitals? Which of the following molecules contain one or more pi bonds? (Select all that apply.) From left to right, give the hybridization of each carbon atom in the allene molecule, H_2C=C=CH_2. Which of the following pairs of atomic orbitals on adjacent nuclei can overlap to form a pi bond? Consider the x axis to be the internuclear axis. Calculate the bond order of N_2^2+, and determine whether it is paramagnetic or diamagnetic. Calculate the bond order of He_2^+. Which if any of the following species has a bond order of 0? Which of the following molecules contain one or more delocalized pi bonds? (Select all that apply.) Which of the atoms in BCl_3, need hybrid orbitals to describe the bonding in the molecule? Which of the following can hybrid orbitals be used for? (Select all that apply.)

Explanation / Answer

13) due to presence of lone pair of electrons the bond angle will be less than tetrahedral angel i.e 109.50

option C

14) Due to bent shape SO2 is polar

Due to pyramidal shape NH3 is polar

Due to linear shape XeF2 is non polar

Answer : E

15) Berylium : Be : 1s2 2s2

Neon : an inert gas

16) configuration of nitorgen is

1s2 2s2 2p3

so : Three bonds

17) The hybrid orbitals = number of atomic orbtials pariticipated

So answer is 1s + 2p = 3 sp2 (three)

18) as clear sp3 , it contans three p orbitals

19) N2 : It contains two pi bonds (triple bond is present between to nitrogens 1 sigma and two pi)

Cl2 : one single bond

CO2 : two pi bonds O=C=O

CH3OH : all single bonds

CCl4 : all single bonds

20 )the hybridisation is

sp2-sp-sp2

21) the p orbitals can form pi bond that too on same axix

So answer is

2py -2py

2px-3px

22) The bond order = Bonding electrons - Antibonding electrons / 2

N2 : 1s2 *1s2 2s2 *2s2 2px2 2py2 2pz2

N2+2 : 1s2 *1s2 2s2 *2s2 2px2 2py2

Bonding electrons = 8

Antibonding electrons = 4

BO = 8-4/2 = 2

23) the electronic configuration of He2+

1s2 *1s1

Bond order = 2-1 /2 = 0.5

24) Bond order zero means , The number of bonding electrons = Number of antibonding electrons

He2+2

25) delocalized pi bond is present in NO2+

26) Only boron , as it is the cental atom of the molecule

27) Both , geometry and bonding in the cental atom

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