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Chapter 4: Problem 43

If the consecutive ionization energies of an element A are \(496,4564,6918,9542 \mathrm{~kJ} / \mathrm{mole}\) respectively, the formula of the oxide formed by \(\mathrm{A}\) is (a) \(\mathrm{A}_{2} \mathrm{O}_{3}\) (b) \(\mathrm{AO}\) (c) \(\mathrm{AO}_{2}\) (d) \(\mathrm{A}_{2} \mathrm{O}\)

Short Answer

Expert verified
Answer: The formula of the oxide formed by element A is A₂O.

Step by step solution

01

Analyze the ionization energies

Look at the given ionization energies. They are in increasing order, which is expected since ionization energies increase as an atom loses more electrons. We are interested in the significant jump between ionization energies, which indicates the change from removing valence electrons to core electrons.
02

Find the jump in ionization energies

Identify the jump in ionization energies by comparing consecutive values: 1st to 2nd: \(\frac{4564-496}{496} \approx 8.2\) 2nd to 3rd: \(\frac{6918-4564}{4564} \approx 0.51\) 3rd to 4th: \(\frac{9542-6918}{6918} \approx 0.38\) The significant jump is between the 1st and 2nd ionization energies.
03

Determine the group of the element

Since the significant jump occurs after the 1st ionization energy, it means that element A has one valence electron. This places element A in Group 1 of the periodic table.
04

Predict the valency and oxide formula

Group 1 elements have a valency of +1 as they tend to lose one electron to achieve stable electronic configuration. Therefore, the valency of element A is +1. Oxygen has a valency of -2. To form a neutral compound, the formula of the oxide will be A_2O.
05

Choose the correct option

Based on our analysis, the correct option is: (d) \(\mathrm{A}_{2} \mathrm{O}\)

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Most popular questions from this chapter

Which one of the following electronic configuration corresponds to the element with maximum electropositive character? (a) \([\mathrm{Kr}] 5 \mathrm{~s}^{1}\) (b) \([\mathrm{Ne}] 3 \mathrm{~s}^{1}\) (c) [Ar]4s (d) [Xe]6s

Which of the following elements acts as the best reducing agent? (a) \(\mathrm{Na}\) (b) \(\mathrm{Cl}\) (c) \(\mathrm{Mg}\) (d) \(\mathrm{F}\)

Column A \(\quad\) Column \(\mathbf{B}\) A. s-block () a. Radioactive metal B. Fluorine () b. Highest first IP value C. Chlorine () c. Lanthanides D. d-block () d. Highly reactive metals E. 4 f-series () e. Heavy metals E Helium () f. Always form \(-1\) oxidation state G. Francium ( ) g. Highest electron affinity value H. Noble gases () \(\mathrm{h}\). Zero group elements

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Transition metals can exhibit variable valency. It is because of (a) the smaller atomic radius (b) the high screening effect (c) the very less energy difference between \((\mathrm{n}-1)\) d-subshell and ns-subshell (d) the high nuclear charge
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