Review concepts of reduction potentials, stability of oxidation states in d-block elements, and the relationship between unpaired electrons and magnetic moment
The reduction potential for the M3+/M2+ couple for manganese is greater than that for iron:
\[ E^\circ_{\text{Mn}^{3+}/\text{Mn}^{2+}} = +1.57 \, \text{V}, \, E^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}} = +0.77 \, \text{V} \]
Therefore, this statement is incorrect.
Higher oxidation states of first-row d-block elements are stabilized by oxide ions (O2−) due to the formation of strong metal-oxygen bonds. This statement is correct.
Chromium in the Cr2+ oxidation state can reduce H+ to H2 in aqueous solution:
\[ \text{Cr}^{2+} + \text{H}^+ \rightarrow \text{Cr}^{3+} + \frac{1}{2}\text{H}_2 \]
The reduction potential \( E^\circ_{\text{Cr}^{3+}/\text{Cr}^{2+}} = -0.26 \, \text{V} \) confirms this. This statement is correct.
V2+ has three unpaired electrons, resulting in a magnetic moment of approximately 3.87 BM, which is not within the range of 4.4-5.2 BM. This statement is incorrect.

If the system of equations \[ (\lambda - 1)x + (\lambda - 4)y + \lambda z = 5 \] \[ \lambda x + (\lambda - 1)y + (\lambda - 4)z = 7 \] \[ (\lambda + 1)x + (\lambda + 2)y - (\lambda + 2)z = 9 \] has infinitely many solutions, then \( \lambda^2 + \lambda \) is equal to:
The output of the circuit is low (zero) for:

(A) \( X = 0, Y = 0 \)
(B) \( X = 0, Y = 1 \)
(C) \( X = 1, Y = 0 \)
(D) \( X = 1, Y = 1 \)
Choose the correct answer from the options given below:
The metal ions that have the calculated spin only magnetic moment value of 4.9 B.M. are
A. $ Cr^{2+} $
B. $ Fe^{2+} $
C. $ Fe^{3+} $
D. $ Co^{2+} $
E. $ Mn^{2+} $
Choose the correct answer from the options given below
Which of the following circuits has the same output as that of the given circuit?
