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Chemical Bonding & Molecular Structure

NEET Chemistry · 95 questions · Page 5 of 10 · Click an option or "Show Solution" to reveal answer

Q41
The pair of species that has the same bond order in the following is
A CO, NO +
B NO −- , CN −-
C O 2 , N 2
D O 2 , B 2
Correct Answer
Option A
Solution

CO = + 8 = 14 electrons NO + = 7 + 8 - 1 = 14 electrons Both have electronic configuration:

σ1s2\sigma 1s^2
σ∗1s2\sigma^* 1s^2
σ2s2\sigma 2s^2
σ∗2s2\sigma^* 2s^2
σ2pz2\sigma 2p_z^2
σ2px2\sigma 2p_x^2
σ2py2\sigma 2p_y^2

Bond order =

10−4210 - 4 \over 2

= 3

Q42
In which of the following ionization processes the bond energy increases and the magnetic behaviour changes from paramagnetic to diamagnetic.
A O 2 → \to O 2 +
B C 2 → \to C 2 +
C NO → \to NO +
D N 2 → \to N 2 +
Correct Answer
Option C
Solution

Molecular orbital configuration of O 2 + ⇒\Rightarrow

σ1s2σ∗1s2σ2s2σ∗2s2σ2pz2π2py2π∗2px1π∗2py1\sigma 1{s^2}{\sigma ^*}1{s^2}\sigma 2{s^2}{\sigma ^*}2{s^2}\sigma 2p_z^2\pi 2p_y^2{\pi ^*}2p_x^1{\pi ^*}2p_y^1

⇒\Rightarrow Paramagnetic Bond order =

10−52=2.5{{10 - 5} \over 2} = 2.5

C 2 ⇒\Rightarrow

σ1s2σ∗1s2σ2s2σ∗2s2σ2pz2π2py2\sigma 1{s^2}{\sigma ^*}1{s^2}\sigma 2{s^2}{\sigma ^*}2{s^2}\sigma 2p_z^2\pi 2p_y^2

⇒\Rightarrow Diamagnetic Bond order =

8−42=2{{8 - 4} \over 2} = 2

C 2 + ⇒\Rightarrow

σ1s2σ∗1s2σ2s2σ∗2s2σ2pz2π2py1\sigma 1{s^2}{\sigma ^*}1{s^2}\sigma 2{s^2}{\sigma ^*}2{s^2}\sigma 2p_z^2\pi 2p_y^1

⇒\Rightarrow Paramagnetic Bond order =

7−42=1.5{{7 - 4} \over 2} = 1.5

NO ⇒\Rightarrow

σ1s2σ∗1s2σ2s2σ∗2s2σ2pz2π2py2π∗2px1\sigma 1{s^2}{\sigma ^*}1{s^2}\sigma 2{s^2}{\sigma ^*}2{s^2}\sigma 2p_z^2\pi 2p_y^2{\pi ^*}2p_x^1

⇒\Rightarrow Paramagnetic Bond order =

10−52=2.5{{10 - 5} \over 2} = 2.5

NO + ⇒\Rightarrow

σ1s2σ∗1s2σ2s2σ∗2s2σ2pz2π2px2π2py2\sigma 1{s^2}{\sigma ^*}1{s^2}\sigma 2{s^2}{\sigma ^*}2{s^2}\sigma 2p_z^2\pi 2p_x^2\pi 2p_y^2

⇒\Rightarrow Diamagnetic Bond order =

10−42=3{{10 - 4} \over 2} = 3

N 2 ⇒\Rightarrow

σ1s2σ∗1s2σ2s2σ∗2s2π2px2π2py2σ2pz2\sigma 1{s^2}{\sigma ^*}1{s^2}\sigma 2{s^2}{\sigma ^*}2{s^2}\pi 2p_x^2\pi 2p_y^2\sigma 2p_z^2

⇒\Rightarrow Paramagnetic Bond order =

10−42=3{{10 - 4} \over 2} = 3

N 2 + ⇒\Rightarrow

σ1s2σ∗1s2σ2s2σ∗2s2π2px2π2py2σ2pz1\sigma 1{s^2}{\sigma ^*}1{s^2}\sigma 2{s^2}{\sigma ^*}2{s^2}\pi 2p_x^2\pi 2p_y^2\sigma 2p_z^1

⇒\Rightarrow Paramagnetic Bond order =

9−42=2.5{{9 - 4} \over 2} = 2.5
Q43
In which of the following pair both the species have sp 3 hybridization ?
A SiF 4 , BeH 2
B NF 3 , H 2 O
C NF 3 , BF 3
D H 2 S, BF 3
Correct Answer
Option B
Solution

NF 3 and H 2 O are sp 3 -hybridisation

Q44
Which of the following is a polar molecule?
A SiF 4
B XeF 4
C BF 3
D SF 4
Correct Answer
Option D
Solution

The given molecules SiF 4 , XeF 4 , and BF 3 are symmetric molecules that is why due to cancellation of bond dipoles they are non-polar even though they contain polar bonds.

But in SiF 4 there are four Si—F bonds and one lone pair due to which its structure is unsymmetrical.

Hence, it is a polar molecule.

Q45
Dipole-induced dipole interactions are present in which of the following pairs
A HCl and He atoms
B SiF 4 and He atoms
C H 2 O and alcohol
D Cl 2 and CCl 4
Correct Answer
Option A
Solution

HCl is polar (μ\mu ≠\ne 0) and He is non-polar (μ\mu = 0) gives dipole-induced dipole interaction.

Q46
Identify the correct order of solubility in aqueous medium.
A Na 2 S > CuS > ZnS
B Na 2 S > ZnS > CuS
C Cus > ZnS > Na 2 S
D ZnS > Na 2 S > CuS
Correct Answer
Option B
Solution

Water is a polar compound so the salt which is more polar or having more ionic character will be more soluble in it.

According to Fajan's rule, ionic character of compound increases with increase in the size of cation.

Now, among the given compounds, the size of cations are in the order as follows : Na + > Zn 2+ > Cu 2+ Thus, the order of ionic character and the solubility order in aqueous medium is as follows : Na 2 S > ZnS > CuS

Q47
Which of the following is paramagnetic?
A CN −-
B NO +
C CO
D O 2 −-
Correct Answer
Option D
Solution
O2−O_2^-

superoxide has one unpaired electron.

Q48
Which of the following is electron-deficient ?
A (BH 3 ) 2
B PH 3
C (CH 3 ) 2
D (SiH 3 ) 2
Correct Answer
Option A
Solution

Boron hydrides are electron deficient compound because boron atom has three half-filled orbitals in excited state.

Q49
During change of O 2 to O2−_2^{ - } ion, the electron adds on which one of the following orbitals ?
A π∗{\pi ^ * } orbital
B π\pi orbital
C σ∗{\sigma ^ * } orbital
D σ\sigma orbital
Correct Answer
Option A
Solution

Electronic configuration of O 2

σ1s2σ∗1s2σ2s2σ∗2s2σ2pz2π2py2π∗2px1π∗2py1\sigma 1{s^2}{\sigma ^*}1{s^2}\sigma 2{s^2}{\sigma ^*}2{s^2}\sigma 2p_z^2\pi 2p_y^2{\pi ^*}2p_x^1{\pi ^*}2p_y^1

Thus the incoming electron will enter in

π∗2px\pi ^*2p_x

to form

O2+O_2^+
Q50
Four diatomic species are listed below. Identify the correct order in which the bond order is increasing in them
A NO<O2−<C22−<He2+NO < {O_2}^ - < {C_2}^{2 - } < He_2^ +
B O2−<NO<C22−<He2+{O_2}^ - < NO < {C_2}^{2 - } < He_2^ +
C C22−<He2+<O2−<NO{C_2}^{2 - } < He_2^ + < {O_2}^ - < NO
D He2+<O2−<NO<C22−He_2^ + < {O_2}^ - < NO < {C_2}^{2 - }
Correct Answer
Option D
Solution

<table class=tg> <tbody><tr> <th class=tg-hgcj>Diatomic species</th> <th class=tg-amwm>Bond order</th> </tr> <tr> <td class=tg-baqh>NO</td> <td class=tg-baqh>2.5</td> </tr> <tr> <td class=tg-baqh>

O2−O_2^-

</td> <td class=tg-baqh>1.5</td> </tr> <tr> <td class=tg-baqh>

C22−C_2^{2-}

</td> <td class=tg-baqh>3.0</td> </tr> <tr> <td class=tg-baqh>

He2+He_2^+

</td> <td class=tg-baqh>0.5</td> </tr> </tbody></table> The increasing order:

He2+<O2−<NO<C22−He_2^ + < {O_2}^ - < NO < {C_2}^{2 - }
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