Read the following passage and answer the questions based on it.

The transition metals are very hard and have low volatility. Their melting and boiling points are high. In any row, the melting points of these metals rise to a maximum at $d^5$ and fall regularly as atomic number increases. The high melting points of these metals are attributed to the involvement of greater number of electrons from (n-1)d in addition to ns electrons in the interatomic metallic bonding.

How many electrons are needed in reduction of $Cr_2O_7^{2-}$ to $Cr^{3+}$?

Six

The correct answer is Option (2) → Six.

To determine how many electrons are needed to reduce \(\text{Cr}_2\text{O}_7^{2-}\) (the dichromate ion) to \(\text{Cr}^{3+}\), we can follow these steps:

Step 1: Identify the Oxidation States

In \(\text{Cr}_2\text{O}_7^{2-}\):

Each oxygen (\(O\)) has an oxidation state of \(-2\).

There are 7 oxygen atoms, contributing a total of \(-14\) (7 × \(-2\)).

Let \(x\) be the oxidation state of chromium (\(Cr\)):

\(2x + (-14) = -2 \implies 2x = 12 \implies x = +6\)

Therefore, each chromium in \(\text{Cr}_2\text{O}_7^{2-}\) is in the +6 oxidation state.

In \(\text{Cr}^{3+}\):

The oxidation state of chromium is \(+3\).

Step 2: Determine the Change in Oxidation State

Change per Chromium Atom:

From +6 to +3, each chromium atom is reduced by \(3\) units.

Total Change for Two Chromium Atoms:

Since there are 2 chromium atoms in \(\text{Cr}_2\text{O}_7^{2-}\):

\(2 \times (6 - 3) = 6 \text{ units}\)

Step 3: Electrons Needed for Reduction

Each unit decrease in oxidation state corresponds to the gain of one electron.

Therefore, to reduce \(\text{Cr}_2\text{O}_7^{2-}\) to \(\text{Cr}^{3+}\), a total of \(6\) electrons are required.

Conclusion

The number of electrons needed in the reduction of \(\text{Cr}_2\text{O}_7^{2-}\) to \(\text{Cr}^{3+}\) is: Six