Polarization of electrons in acrolein may be written as: 

1. $\stackrel{{\delta }^{-}}{C}{H}_2=CH-\stackrel{{\delta }^{+}}{C}H=O$

2. $\stackrel{{\delta }^{-}}{C}{H}_2=CH-CH=\stackrel{{\delta }^{+}}{O}$

3. $\stackrel{{\delta }^{-}}{C}{H}_2=\stackrel{{\delta }^{+}}{C}H-CH=O$

4. $\stackrel{{\delta }^{+}}{C}{H}_2=CH-CH=\stackrel{{\delta }^{-}}{O}$

$PC{l}_5$ exists but $NC{l}_5$ does not because:

1. Nitrogen has no vacant d-subshells .

2. $NC{l}_5$ is unstable .

3. Nitrogen atom is much smaller than p .

4. Nitrogen is highly inert . 

The bond angle between two hybrid orbitals is ${105}^{°}$. The % s character in orbital is-

Which bond angle would result in the maximum dipole moment for the triatomic molecule $X{Y}_2$ 

1. ${90}^{°}$

2. ${120}^{°}$

3. ${150}^{°}$

4. ${180}^{°}$

The strongest hydrogen bond present among the following is-

1. $\mathrm{O}—\mathrm{H}‐‐‐\mathrm{S}$

2. $\mathrm{S}—\mathrm{H}‐‐‐\mathrm{O}$

3. $\mathrm{F}—\mathrm{H}‐‐‐\mathrm{F}$

4. $\mathrm{F}—\mathrm{H}‐‐‐\mathrm{O}$

In which of the process, the bond order increases and magnetic behaviour changes?

(a) $N_2\to {N_2}^{+}$

(b) $C_2\to {C_2}^{+}$

(c) $NO\to N{O}^{+}$

(d) $O_2\to {O_2}^{+}$

In piperidine "N" atom has hybridization:

1. sp

2. $s{p}^2$

3. $s{p}^3$

4. $ds{p}^2$

The coordination geometry around and hybridization of N and B atoms in a 1:1 complex of NH₃ and BF₃ ${\mathrm{is}}_{}$:

(1) N: tetrahedral, sp³;  B: tetrahedral, sp³

(2) N: pyramidal, sp³; B: pyramidal, sp³

(3) N: pyramidal, sp³; B: planar, sp²

(4) N: pyramidal, sp³; B: tetrahedral, sp³

Assuming $2s-2p$ mixing is NOT operative, the paramagnetic species among the following is :

(1) $B{e}_2$

(2) $B_2$

(3) $C_2$

(4) $N_2$

When $O_2$ is adsorbed on a metallic surface, electron transfer occurs from the ,metal to $O_2$. The wrong statements(s) regarding this adsorption is (are): 

(1) $O_2$ is physisorbed 

(2) Heat is released

(3) Occupancy of $\pi *2p$ of $O_2$ is increased 

(4) Bond length of $O_2$ is increased