Match List I with List II
Choose the correct answer from the options given below:
Match List I with List II
Choose the correct answer from the options given below:
- A - IV, B - I, C - II, D - III
- A - II, B - I, C - IV, D - III
- A - II, B - III, C - IV, D - I
- A - IV, B - III, C - II, D - I
The Correct Option is D
Solution and Explanation
Step 1: Determine the Lone Pairs for Each Molecule/Ion
The number of lone pairs on the central atom can be determined using the following steps:
Count the total valence electrons of the central atom.
Subtract the electrons used for bonding with surrounding atoms.
Divide the remaining electrons by 2 to get the number of lone pairs.
Step 2: Analyze Each Molecule/Ion
\(\text{IF}_7\): Iodine has 7 valence electrons. All are used for bonding with 7 fluorine atoms. Therefore, 0 lone pairs (IV).
\(\text{ICl}_4^-\): Iodine has 7 valence electrons and gains 1 due to the negative charge. Four are used for bonding with chlorine atoms, leaving 4 electrons (2 lone pairs). Therefore, 2 lone pairs (III).
\(\text{XeF}_6\): Xenon has 8 valence electrons. Six are used for bonding with fluorine atoms, leaving 2 electrons (1 lone pair). Therefore, 1 lone pair (II).
\(\text{XeF}_2\): Xenon has 8 valence electrons. Two are used for bonding with fluorine atoms, leaving 6 electrons (3 lone pairs). Therefore, 3 lone pairs (I)}
Top JEE Main Chemistry Questions
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Top JEE Main p -Block Elements Questions
- $\textbf{Reduction potential of ions are given below:}$
\[ \begin{array}{ccc} \text{ClO}_4^- & \text{IO}_4^- & \text{BrO}_4^- \\ E^\circ = 1.19 \, \text{V} & E^\circ = 1.65 \, \text{V} & E^\circ = 1.74 \, \text{V} \\ \end{array} \]
The correct order of their oxidising power is: - $\textbf{Choose the correct statements about the hydrides of group 15 elements.}$
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 \(\text{NH}_3 < \text{PH}_3 < \text{AsH}_3 < \text{SbH}_3 < \text{BiH}_3\)
Choose the most appropriate from the option given below: The number of oxygen atoms present in chemical formula of fuming sulphuric acid is _______.
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(B) Non metals have lower ionisation enthalpy than metals.
(C) Compounds formed between highly reactive nonmetals and highly reactive metals are generally ionic.
(D) The non-metal oxides are generally basic in nature.
(E) The metal oxides are generally acidic or neutral in nature. 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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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