1
JEE Main 2023 (Online) 31st January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Match List I with List II

List I List II
A. $$\mathrm{XeF_4}$$ I. See-saw
B. $$\mathrm{SF_4}$$ II. Square-planar
C. $$\mathrm{NH_{4}^{+}}$$ III. Bent T-shaped
D. $$\mathrm{BrF_3}$$ IV. Tetrahedral

Choose the correct answer from the options given below :

A
A - II, B - I, C - III, D - IV
B
A - II, B - I, C - IV, D - III
C
A - IV, B - I, C - II, D - III
D
A - IV, B - III, C - II, D - I
2
JEE Main 2023 (Online) 31st January Morning Shift
Numerical
+4
-1
Change Language

Zinc reacts with hydrochloric acid to give hydrogen and zinc chloride. The volume of hydrogen gas produced at STP from the reaction of $$11.5 \mathrm{~g}$$ of zinc with excess $$\mathrm{HCl}$$ is __________ L (Nearest integer)

(Given : Molar mass of $$\mathrm{Zn}$$ is $$65.4 \mathrm{~g} \mathrm{~mol}^{-1}$$ and Molar volume of $$\mathrm{H}_{2}$$ at $$\mathrm{STP}=22.7 \mathrm{~L}$$ )

Your input ____
3
JEE Main 2023 (Online) 31st January Morning Shift
Numerical
+4
-1
Change Language

The logarithm of equilibrium constant for the reaction $$\mathrm{Pd}^{2+}+4 \mathrm{Cl}^{-} \rightleftharpoons \mathrm{PdCl}_{4}^{2-}$$ is ___________ (Nearest integer)

Given : $$\frac{2.303 R \mathrm{~T}}{\mathrm{~F}}=0.06 \mathrm{~V}$$

$$ \mathrm{Pd}_{(\mathrm{aq})}^{2+}+2 \mathrm{e}^{-} \rightleftharpoons \mathrm{Pd}(\mathrm{s}) \quad \mathrm{E}^{\ominus}=0.83 \mathrm{~V} $$

$$ \begin{aligned} & \mathrm{PdCl}_{4}^{2-}(\mathrm{aq})+2 \mathrm{e}^{-} \rightleftharpoons \mathrm{Pd}(\mathrm{s})+4 \mathrm{Cl}^{-}(\mathrm{aq}) \mathrm{E}^{\ominus}=0.65 \mathrm{~V} \end{aligned} $$

Your input ____
4
JEE Main 2023 (Online) 31st January Morning Shift
Numerical
+4
-1
Change Language

The enthalpy change for the conversion of $$\frac{1}{2} \mathrm{Cl}_{2}(\mathrm{~g})$$ to $$\mathrm{Cl}^{-}$$(aq) is ($$-$$) ___________ $$\mathrm{kJ} \mathrm{mol}^{-1}$$ (Nearest integer)

Given : $$\Delta_{\mathrm{dis}} \mathrm{H}_{\mathrm{Cl}_{2(\mathrm{~g})}^{\theta}}^{\ominus}=240 \mathrm{~kJ} \mathrm{~mol}^{-1}, \Delta_{\mathrm{eg}} \mathrm{H}_{\mathrm{Cl_{(g)}}}^{\ominus}=-350 \mathrm{~kJ} \mathrm{~mol}^{-1}$$,

$${\mathrm{\Delta _{hyd}}H_{Cl_{(g)}^ - }^\Theta = - 380}$$ $$\mathrm{kJ~mol^{-1}}$$

Your input ____
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