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Important Questions on Chemical Bonding
1. Predict the shape of PCl4+ ion using VSEPR theory.
- Trigonal planar
2. Predict the shape of the XeF2 molecule.
- Trigonal bipyramidal
- Square planar
3. The correct order of O-O bond length in O2, H2O2 and O3 is:
- O3>O2> H2O2
4. The formation of molecular orbitals from atomic orbitals causes electron density to be
- Zero in nodal plane
- Zero in the surface of the lobe
- Minimum in nodal plane
- Maximum in nodal plane
5. Incorrect order of bond length is:
- CO < CO+
- NO > NO+
- N2+ < N2–
6. Order of increasing stability is:
- N22- < N2- = N2+ < N2
- N22- < N2- < N2+ < N2
- N22- = N2+ < N2- < N2
- N22- < N2+< N2- < N2
7. The paramagnetic species are:
8. The main axis of a diatomic molecule is Z. The orbitals Px and Py overlap to form
- - molecular orbital
- – molecular orbital
- 𝛿 - molecular orbital
- No bond is formed.
9. How many nodal surfaces can be present in the molecular orbital formed by the overlap of s and px (x-axis is the molecular axis)?
- None of the above
10. The dipole moment of certain diatomic molecule X – Y is 0.38 D . If the X – Y distance is 158 pm, the percentage of electronic charge developed on X-atoms is:
P contains five valence electrons. Out of these five electrons, four will be used in bonding with four Cl atoms. There is a positive charge which indicates the removal of one electron. In this, there is no lone pair. So, its shape is tetrahedral.
XeF2= 1/2(8+2) =5
As 2 F are present, they will be used in bonding. There will be three lone pairs that will occupy equatorial positions.
So, it is linear in shape.
When electrons are present in bonding orbitals, they spend most of the time between the nuclei of two atoms while when they are present in antibonding orbitals, they spend their time outside the nucleus. This is because there is an increase in electron density in bonding and a decrease in electron density in antibonding. The amplitude of the wave function is zero, and hence the probability of finding an electron in the region.
The Molecular orbital configuration of CO is:
Removal of one electron from CO will result in an increase in bond order. Hence, the bond order of CO < CO+
Therefore, the bond length order will be:
CO > CO+.
The order of stability is directly proportional to the bond order. Therefore, the correct order of stability is N22- < N2– = N2+ < N2
KO2→K+ + O2- =17e- in valence shell so unpaired electron present and shows paramagnetism.
Rest all have O22- dioxide ion in which all the electrons are paired and diamagnetic.
No bond is formed as no overlap is possible between Px and Py.
% Electron charge is:
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