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Chapter 4: Problem 36

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

Short Answer

Expert verified
(a) [Kr]5s1 (b) [Ne]3s1 (c) [Ar]4s2 (d) [Xe]6s2 Answer: (d) [Xe]6s2 (Barium)

Step by step solution

01

Identify the elements for each electronic configuration

We will first identify the elements corresponding to the given electronic configurations. (a) \([\mathrm{Kr}] 5 \mathrm{~s}^{1}\) corresponds to the element with atomic number 36 + 1 = 37, which is Rubidium (Rb) (b) \([\mathrm{Ne}] 3 \mathrm{~s}^{1}\) corresponds to the element with atomic number 10 + 1 = 11, which is Sodium (Na) (c) [Ar]4s corresponds to the element with atomic number 18 + 2 = 20, which is Calcium (Ca) (d) [Xe]6s corresponds to the element with atomic number 54 + 2 = 56, which is Barium (Ba)
02

Locate the elements in the periodic table

Locate the elements Rubidium (Rb), Sodium (Na), Calcium (Ca), and Barium (Ba) in the periodic table. We will find that they are in order as follows: - Sodium (Na) is in Group 1 and Period 3 - Rubidium (Rb) is in Group 1 and Period 5 - Calcium (Ca) is in Group 2 and Period 4 - Barium (Ba) is in Group 2 and Period 6
03

Compare the electropositive character

Recall that electropositive character increases down a group and decreases across a period. Since Rb and Ba are in the same group but Ba is in a lower period, Ba will have a stronger electropositive character than Rb. Na is in the same period as Rb but with a lower group number, meaning its electropositive character is weaker than that of Rb. Ca is in the same group as Ba but with a higher period, meaning its electropositive character will be weaker than Ba. Comparing these trends, we can conclude that the element with the maximum electropositive character among the given electronic configurations is Barium (Ba) with the electronic configuration [Xe]6s. The correct answer is (d).

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Most popular questions from this chapter

The total number of elements present in the 6 th period is (a) 18 (b) 31 (c) 32 (d) 17

Fill in the blanks. In most of the lanthanides, the penultimate shell \(\begin{array}{lll}\text { contains } & \text { electrons in } & \text { orbitals. }\end{array}\)

An element with atomic number " \(32 "\) "belongs to (a) \(4^{\text {th }}\) period, VIA group or \(16^{\text {th }}\) group (b) \(3^{\text {rd }}\) period, IVA group or \(14^{\text {th }}\) group (c) \(4^{\text {th }}\) period, IVA group or \(14^{\text {th }}\) group (d) \(5^{\text {th }}\) period, VA group or \(15^{\text {th }}\) group

Which of the following sequence of explanation is appropriate for explaining the reason for the periodicity of reducing property in a period or group? (1) Tendency to undergo oxidation decreases in a period and increases in a group. (2) In a period ionization energy increases and in a group it decreases. (3) In a period the atomic size decreases and in a group it increases. (4) The elements present in the left side of the periodic table have strong reducing property. (a) 3214 (b) 3421 (c) 4312 (d) 4213

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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