MCQ
Incorrect statement regarding $BF_2NH_2$ molecule is
- A$FBF$ bond angle $< 120^o$
- B$HNH$ bond angle $> 109^o28'$
- CExlubits intermolecular $H-$ bond
- ✓Hybridization of $N-$ atom is $sp^3$
$*\, FBF$ bond angle $< 120^o$ ($VSPER$ theory)
$* \,H N H$ bond angle is less than $120^o$ but greater than $109^o28$' due to back bonding.
$*$ Due to presence of $H-$ atom attached to nitrogen this molecule can exhibits intermolecular $H-$ bonding.
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| Rate constant | Activation energy | |
| Step $1$ | $k_1$ | $E_{a1} = 180\ kJ/mol$ |
| Step $2$ | $k_2$ | $E_{a2} = 80\ kJ/mol$ |
| Step $3$ | $k_3$ | $E_{a3} = 50\ kJ/mol$ |
If overall rate constant, $k = {\left( {\frac{{{k_1}{k_2}}}{{{k_3}}}} \right)^{2/3}}$ ,then overall activation energy of the reaction will be .......... $ kJ/mol$
Reason : Iodine is a polar compound.
| $I$ | $II$ | |
| $[Cr(H_2O)_6]^{+2}$ | $[Cr(H_2O)_6]^{+3}$ | $(a)$ |
| $[Fe(H_2O)_6]^{+3}$ | $[Fe(CN)_6]^{-3}$ | $(b)$ |
| $[Fe(CN)_6]^{+3}$ | $[Ru(CN)_6]^{-3}$ | $(c)$ |
| $[NiF_6]^{-4}$ | $[NiF_6]^{-2}$ | $(d)$ |