Chemistry MCQs for NEET — Practice Questions with Answers

In qualitative analysis, the passage of $H_2S$ gas in the presence of dilute HCl is used to detect second group basic radicals (like $Cu^{2+}$, $Pb^{2+}$, etc.). The presence of HCl decreases the dissociation of $H_2S$ due to the common ion effect. This ensures that only the second group radicals, which form insoluble sulfides, precipitate out.

$$ H_2S ightleftharpoons 2H^+ + S^{2-} $$

The addition of HCl increases the concentration of $H^+$ ions, shifting the equilibrium to the left and thus decreasing the dissociation of $H_2S$.

Sodium nitroprusside reacts with sulphide ions in an alkaline solution to produce a purple color. The reaction involved is:

$$ [Fe(CN)_5NO]^{2-} + S^{2-} ightarrow [Fe(CN)_5NOS]^{4-} $$

This complex ion $[Fe(CN)_5NOS]^{4-}$ is responsible for the purple coloration observed in the solution.

To determine the nature of the resulting solution, we must first calculate the milliequivalents of the acid and the base. For the acid: \[20 \, \text{ml} \times 0.25 \, N = 5 \, \text{milliequivalents}\]. For the base: \[30 \, \text{ml} \times 0.2 \, N = 6 \, \text{milliequivalents}\]. Since the base has more milliequivalents, the solution will be basic. To find the normality of the resulting solution: \[\text{Excess milliequivalents of base} = 6 - 5 = 1 \, \text{milliequivalent}\] and the total volume is \[20 + 30 = 50 \, \text{ml}\]. Thus, the normality is \[\frac{1}{50/1000} = 0.02 \, N\]. Since the base is in excess, the resulting solution is 0.02 N basic.

To determine the nature of the resulting solution, we calculate the number of millimoles of $H_2SO_4$ and NaOH. For $H_2SO_4$: \[10 \, \text{ml} \times 10 \, \text{M} \times 2 = 200 \text{millimoles of } H^+\] (since $H_2SO_4$ is diprotic). For NaOH: \[100 \, \text{ml} \times 1 \, \text{M} = 100 \text{millimoles of } OH^-\]. Since the millimoles of $H^+$ are greater than the millimoles of $OH^-$, the solution will be acidic.

To solve this, we need to use the concept of equivalent weights and normality. The equivalent weight of oxalic acid ($C_2H_2O_4 imes 2H_2O$) is calculated as follows: Molecular weight of oxalic acid dihydrate = 126 g/mol. Since oxalic acid is diprotic (gives 2 H+ ions), its equivalent weight is 126/2 = 63 g/equiv. Therefore, the weight of oxalic acid crystals in 1N solution will be 63g.

The most electronegative element is fluorine, which has the electronic configuration $1s^2 2s^2 2p^5$. This matches the general form $ns^2 np^5$, where n=2 for fluorine. Therefore, the correct option is $ns^2 np^5$.

First, we need to convert the mass of Cs to moles: \[ \text{Moles of Cs} = \frac{\text{mass}}{\text{molar mass}} = \frac{2.66 \text{ mg}}{133 \text{ g/mol}} \] Since 1 mg = 0.001 g, \[ \text{Moles of Cs} = \frac{2.66 \times 10^{-3} \text{ g}}{133 \text{ g/mol}} \approx 2 \times 10^{-5} \text{mol} \] The ionization enthalpy of Cs is given as 375.6 kJ/mol. Therefore, the energy required to convert 2.66 mg of Cs to \( Cs^+ \) is: \[ \text{Energy} = \text{moles} \times \text{ionization enthalpy} = 2 \times 10^{-5} \text{ mol} \times 375.6 \text{ kJ/mol} \] \[ \text{Energy} = 7.512 \text{ J} \] Thus, the correct option is 7.512 J.