List-I (Compound) | List-II (Colour) |
---|---|
(A) \(Fe_4[Fe(CN)_6]_3.xH_2O\) | (I) Violet |
(B) \( [Fe(CN)_5NOS]^{4–} \) | (II) Blood Red |
(C) \( [Fe(SCN)]^{2+}\) | (III) Prussian Blue |
(D) \((NH_4)_3PO_4.12MoO_3\) | (IV) Yellow |
Matching Explanation :
Fe$_4$[Fe(CN)$_6$]$_3$ $\cdot$ $x$H$_2$O: Prussian Blue (Colour: III)
[Fe(CN)$_5$NOS]$^{4-}$: Violet (Colour: I)
[Fe(SCN)]$^{2+}$: Blood Red (Colour: II)
(NH$_4$)$_3$PO$_4$ $\cdot$ 12MoO$_3$: Yellow (Colour: IV)
A-III, B-I, C-II, D-IV
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: \(\text{NH}_3, PhSH, (H_3C_2S)_2, H_2C = CH_2, OH−, H_3O+, (CH_3)_2CO, NCH_3\)