Match List-I with List-II
List-I
List-II
(A) Cd(s)+2Ni(OH)3(s)→CdO(s)+2Ni(OH)2(s)+H2O(l) (I) Primary battery (B) Zn(Hg)+HgO(s)→ZnO(s)+Hg(l) (II) Discharging of secondary battery (C) 2PbSO4(s)+2H2O(l)→Pb(s)+PbO2(s)+2H2SO4(aq) (III) Fuel cell (D) 2H2(g)+O2(g)→2H2O(l) (IV) Charging of
secondary battery
Choose the correct answer from the options given below:
List-I | List-II | ||
|---|---|---|---|
| (A) | Cd(s)+2Ni(OH)3(s)→CdO(s)+2Ni(OH)2(s)+H2O(l) | (I) | Primary battery |
| (B) | Zn(Hg)+HgO(s)→ZnO(s)+Hg(l) | (II) | Discharging of secondary battery |
| (C) | 2PbSO4(s)+2H2O(l)→Pb(s)+PbO2(s)+2H2SO4(aq) | (III) | Fuel cell |
| (D) | 2H2(g)+O2(g)→2H2O(l) | (IV) | Charging of secondary battery |
Choose the correct answer from the options given below:
- (A) - (I), (B) - (II), (C) - (III), (D) - (IV)
- (A) - (IV), (B) - (I), (C) - (II), (D) - (III)
- (A) - (II), (B) - (I), (C) - (IV), (D) - (III)
- (A) - (II), (B) - (I), (C) - (III), (D) - (IV)
The Correct Option is C
Solution and Explanation
From the given options the correct answer is option (C): (A) - (II), (B) - (I), (C) - (IV), (D) - (III)
Top Questions on p -Block Elements
Given below are two statements.
In the light of the above statements, choose the correct answer from the options given below:- JEE Main - 2025
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Given below are two statements:
Statement I: Nitrogen forms oxides with +1 to +5 oxidation states due to the formation of $\mathrm{p} \pi-\mathrm{p} \pi$ bond with oxygen.
Statement II: Nitrogen does not form halides with +5 oxidation state due to the absence of d-orbital in it.
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Given below are the pairs of group 13 elements showing their relation in terms of atomic radius. $(\mathrm{B}<\mathrm{Al}),(\mathrm{Al}<\mathrm{Ga}),(\mathrm{Ga}<\mathrm{In})$ and $(\mathrm{In}<\mathrm{Tl})$ Identify the elements present in the incorrect pair and in that pair find out the element (X) that has higher ionic radius $\left(\mathrm{M}^{3+}\right)$ than the other one. The atomic number of the element (X) is
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A. The stability of the hydrides decreases in the order \(\text{NH}_3 > \text{PH}_3 > \text{AsH}_3 > \text{SbH}_3 > \text{BiH}_3\)
B. The reducing ability of the hydrides increases in the order \(\text{NH}_3 < \text{PH}_3 < \text{AsH}_3 < \text{SbH}_3 < \text{BiH}_3\)
C. Among the hydrides, \(\text{NH}_3\) is a strong reducing agent while \(\text{BiH}_3\) is a mild reducing agent.
D. The basicity of the hydrides increases in the order \(\text{NH}_3 < \text{PH}_3 < \text{AsH}_3 < \text{SbH}_3 < \text{BiH}_3\)
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\[ \begin{array}{ccc} \text{ClO}_4^- & \text{IO}_4^- & \text{BrO}_4^- \\ E^\circ = 1.19 \, \text{V} & E^\circ = 1.65 \, \text{V} & E^\circ = 1.74 \, \text{V} \\ \end{array} \]
The correct order of their oxidising power is:- JEE Main - 2024
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Match List-I with List-II: List-I
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Concepts Used:
P-Block Elements
- P block elements are those in which the last electron enters any of the three p-orbitals of their respective shells. Since a p-subshell has three degenerate p-orbitals each of which can accommodate two electrons, therefore in all there are six groups of p-block elements.
- P block elements are shiny and usually a good conductor of electricity and heat as they have a tendency to lose an electron. You will find some amazing properties of elements in a P-block element like gallium. It’s a metal that can melt in the palm of your hand. Silicon is also one of the most important metalloids of the p-block group as it is an important component of glass.
P block elements consist of:
- Group 13 Elements: Boron family
- Group 14 Elements: Carbon family
- Group 15 Elements: Nitrogen family
- Group 16 Elements: Oxygen family
- Group 17 Elements: Fluorine family
- Group 18 Elements: Neon family