List-I (Molecule) | List-II (Shape) |
---|---|
(A) \(NH_3\) | (I) Square pyramid |
(B) \(BrF_5\) | (II) Tetrahedral |
(C) \(PCL_5\) | (III) Trigonal pyramidal |
(D) \(CH_4\) | (IV) Trigonal bipyramidal |
NH$_3$: Trigonalpyramidal ({sp}$^3$ hybridization with one lone pair on N).
BrF$_5$: Squarepyramidal ({sp}$^3${d}$^2$ hybridization with one lone pair on Br).
PCl$_5$: Trigonalbipyramidal ({sp}$^3${d} hybridization, no lone pairs).
CH$_4$: Tetrahedral ({sp}$^3$ hybridization, no lone pairs).
Matching: A-III, B-I, C-IV, D-II.
Final Answer: A-III, B-I, C-IV, D-II.
Let $ f(x) = \begin{cases} (1+ax)^{1/x} & , x<0 \\1+b & , x = 0 \\\frac{(x+4)^{1/2} - 2}{(x+c)^{1/3} - 2} & , x>0 \end{cases} $ be continuous at x = 0. Then $ e^a bc $ is equal to
Total number of nucleophiles from the following is: \(\text{NH}_3, PhSH, (H_3C_2S)_2, H_2C = CH_2, OH−, H_3O+, (CH_3)_2CO, NCH_3\)