MCQ
In a reaction $2A + B \to {A_2}B$, the reactant $ A $ will disappear at
- AHalf the rate that $B$ will decrease
- BThe same rate that $ B$ will decrease
- ✓Twice the rate that $B$ will decrease
- DThe same rate that ${A_2}B$ will form
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
| List- I | List- II |
| (A) $\left[ Ni ( Cl )_4\right]^{-2}$ | (i) $5.92 B . M$. |
| (B) $\left[ CO \left( C _2 O _4\right)_3\right]^{-3}$ | (ii) $1.73 B . M$. |
| (C) $\left[ FeF _6\right]^{-3}$ | (iii) 0 B.M. |
| (D) $\left[ Mn ( CN )_6\right]^{-4}$ | (iv) 2.82 B.M. |
$(A)$ $CO _2, C _2 H _4, NO$ and $HCl$
$(B)$ $NO _2, O _3, HCl$ and $H _2 SO _4$
$(C)$ $BCl _3, NO , NO _2$ and $H _2 SO _4$
$(D)$ $CO _2, BCl _3, O _3$ and $C _2 H _4$
($A$) $\left[\mathrm{Pt}(\mathrm{en})(\mathrm{SCN})_2\right]$
($B$) $\left[\mathrm{Zn}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2\right]$
($C$) $\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_4\right]$
($D$) $\left[\mathrm{Cr}(\mathrm{en})_2\left(\mathrm{H}_2 \mathrm{O}\right)\left(\mathrm{SO}_4\right)\right]^{+}$
Reason : There is delocalisation of electrons from filled $d$ orbitals into the empty orbitals on the $CO$ ligands.