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Chemical Bonding and Molecular Structure

Multiple Choice Questions

391.

Which one of the following has the shortest bond length?

C - H
C - N
C - F
C - C

392.

Pick out the wrong statement.

Nitrogen has the ability to form p $π$ -p $π$ bonds with itself
Bismuth forms metallic bonds in elemental state
Catenation tendency is higher in nitrogen when compared with other elements of the same group
Nitrogen has higher first ionization enthalpy when compared with other elements of the same group

393.

The wrong statement about the interstitial compounds is

their melting points are lower than those of pure metals
they retain metallic conductivity
they are chemically inert
their bonds are neither ionic non covalent

394.

Which of the following is not correct with respect to bond length of the species?

C₂ > $C_{2}^{2 -}$
$B_{2}^{+}$ > B₂
${Li}_{2}^{+}$ > Li₂
O₂ > O $_{2}^{-}$

395.

Substance which is weakly repelled by a magnetic field is

O₂
H₂O
CrO₂
Fe₃O₄

396.

The bonds present in the structure of dichromate ion are

four equivalent Cr-O bonds only
six equivalent Cr-O bonds and one Cr - O - O bond
six equivalent Cr-O bonds and one O-O bond
eight equivalent Cr-O bonds

397.

Which of the following statements is false?

H₂ molecule has 1σ bond
Acetylene molecule has 3σ bonds and 3Π bonds
HCl molecule has 1σ bond
Water molecule has 2σ bonds and two lone pairs

398.
N₂ and O₂ are converted to monopositive cation $N_{2}^{+}$ and $O_{2}^{+}$ respectively. Which is incorrect?

In $N_{2}^{+}$ , the N-N bond weakened.

In $O_{2}^{+},$ the bond order increases

In $O_{2}^{+},$ paramagnetism decreases

$N_{2}^{+}$ becomes diamagnetic

$N_{2}^{+}$ becomes diamagnetic

$MO configuration of N_{2} molecule is (σ 1 s)^{2} (σ^{*} 1 s)^{2} (σ 2 s)^{2} (σ^{*} 2 s)^{2} (π 2 p_{x})^{2} (π 2 p_{y})^{2} (σ 2 p_{z})^{2} ∴ Bond order of N_{2} = \frac{10 - 4}{2} = 3 MO configuration of N_{2}^{+} is (σ 1 s)^{2} (σ^{*} 1 s)^{2} (σ 2 s)^{2} (σ^{*} 2 s)^{2} (π 2 p_{x})^{2} (π 2 p_{y})^{2} (σ 2 p_{z})^{1} Thus, N_{2}^{+} becomes paramagnetic ∵ bond order of N_{2}^{+} = \frac{9 - 4}{2} = 2.5 and bonder \infty bond energy . ∴ In N_{2}^{+}, the N - N bond is weakened . MO configuration of O_{2} is (σ 1 s)^{2} (σ^{*} 1 s)^{2} (σ 2 s)^{2} (σ^{*} 2 s)^{2} (σ 2 p_{z})^{2} (π 2 p_{x})^{2} (π 2 p_{y})^{2} (π^{*} 2 p_{x})^{1} (π^{*} 2 p_{y})^{1} ∴ bond order = \frac{10 - 6}{2} = 2 MO configuration O_{2}^{+} (σ 1 s)^{2} (σ^{*} 1 s)^{2} (σ 2 s)^{2} (σ^{*} 2 s)^{2} (σ 2 p_{z})^{2} (π 2 p_{x})^{2} (π 2 p_{y})^{2} (π^{*} 2 p_{x})^{1} ∴ bond order = \frac{10 - 5}{2} = 2.5 Hence, O_{2} and O_{2}^{+} both are paramagnetic, but in O_{2}^{+} paramagnetic decreases .$