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Introduction to d and f Block Elements

cr2+ is reduc...

Question

cr2+ is reducing in nature while with the ame d-orbital configuration d4 mn3+ is an oxidising agent.explain

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Solution

Cr is in low oxidation state in Cr²⁺. Therefore, it tends to get oxidized to Cr³⁺ by loosing an electron and achieving stable configuration of half filled t_2g of d orbitals.

Mn is in high oxidation state in Mn³⁺. Therefore, it tends to get reduced to Mn²⁺ by gaining an electron and achieving a stable configuration of half filled d orbitals (d⁵).

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Similar questions

C r^{2 +}

is reducing in nature while the same

d -

orbital configuration

(d^{4}) M n^{+ 3}

is an oxidising agent

C r^{2 +}

is reducing and

M n^{3 +}

is oxidising. Reason:

C r^{2 +}

M n^{3 +}

d^{4}

Q. Explain each of the following observations:
(i) With the same d-orbital configuration

(d^{4}), C r^{2 +}

is a reducing agent while

M n^{3 +}

is an oxidising agent.
(ii) Actinoids exhibit a much larger number of oxidation states than lanthanoids.
(iii) There is hardly any increase in atomic size with increasing atomic numbers in a series of transition metals.

Q. The correct statements (s) about

C r^{2 +}

M n^{3 +}

is (are) [Atomic numbers of Cr = 24 and Mn = 25]

Q. How would you account for the following?
(i) With the same d-orbital configuration

(d^{4}) C r^{2 +}

is reducing agent while

M n^{3 +}

is an oxidizing agent.
(ii) The actinoids exhibit a larger number of oxidation states than the corresponding members in the lanthanoid series.
(iii) Most of the transition metal ions exhibit characteristic in colours in aqueous solutions.

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