Problem 67 - A Complete Resource Book in Chemistry for JEE Main

Question

How many unpaired electrons are present in \(\mathrm{Ni}^{2+}\) ? (a) 8 (b) 4 (c) 2 (d) 0

Step-by-Step Solution

Answer

(c) 2

Step 1: Determine the electronic configuration of Ni atom

Nickel (Ni) has an atomic number of 28. The electron configuration of an atom is determined by distributing electrons across the orbitals following the Aufbau principle, Pauli-exclusion principle, and Hund's rule.So, for a neutral Ni atom, the electron configuration is: \[ ext{Ni: } 1s^2 ext{ } 2s^2 ext{ } 2p^6 ext{ } 3s^2 ext{ } 3p^6 ext{ } 4s^2 ext{ } 3d^8 \]

Step 2: Determine the electronic configuration of Ni^{2+} ion

The \( ext{Ni}^{2+}\) ion is formed by the removal of two electrons from the neutral Ni atom. In transition metals, the electrons are first removed from the 4s orbital before the 3d orbital when forming cations.So, the electron configuration for \( ext{Ni}^{2+}\) is: \[ 1s^2 ext{ } 2s^2 ext{ } 2p^6 ext{ } 3s^2 ext{ } 3p^6 ext{ } 3d^8 \] Here, the two electrons removed are from the 4s orbital (4s^2).

Step 3: Count the number of unpaired electrons

In \( ext{Ni}^{2+}\), we are interested in the 3d electrons:- The 3d^8 configuration means there are 8 electrons in the 3d orbitals.- The placement of these 8 electrons in the five 3d orbitals follows Hund's Rule, meaning electrons will fill each orbital singly first before pairing.- The specific filling will be: \[ \begin{array}{c|c|c|c|c}\uparrow & \uparrow & \uparrow & \downarrow & \downarrow \\uparrow & \downarrow\end{array} \] So, \( ext{Ni}^{2+}\) will have 2 unpaired electrons.

Key Concepts

Electronic Configuration

Understanding electronic configuration is crucial to determining the properties of an atom and its ions, such as the presence of unpaired electrons. Electronic configuration is the distribution of electrons among the various atomic orbitals. It follows a specific order based on increasing energy levels and is governed by a few key principles.

To write the electronic configuration, one must follow:

• The Aufbau Principle: Electrons fill orbitals starting with the lowest energy level and proceed to higher ones.
• Pauli Exclusion Principle: Each orbital can hold a maximum of two electrons with opposite spins.
• Hund's Rule: Electrons fill degenerate orbitals (orbitals of the same energy) singly first with parallel spins before pairing up.

A nickel (Ni) atom, with an atomic number of 28, has the electronic configuration: \[ \text{Ni: } 1s^2 \ 2s^2 \ 2p^6 \ 3s^2 \ 3p^6 \ 4s^2 \ 3d^8 \]This tells us how the electrons are distributed across the various orbitals, which is essential for predicting chemical behavior.

Hund's Rule

Hund's Rule is an essential concept in understanding how electrons are configured within an atom, which is key to determining unpaired electrons. In a set of degenerate orbitals, electrons will occupy empty orbitals singly before starting to pair up.

For example, when looking at the 3d orbitals in \(\text{Ni}^{2+}\), the rule suggests placing one electron per orbital to minimize electron repulsion and produce the highest multiplicity. In the case of \(3d^{8}\), the electron filling diagram according to Hund's Rule would look like this:

• Fill each of the five 3d orbitals with one electron first.
• Add the remaining three electrons following the same principle.

This arrangement results in multiple unpaired electrons, more precisely two unpaired electrons in \(\text{Ni}^{2+}\). Hund's Rule ensures that we can determine this distribution correctly and consistently.

Electron Removal in Transition Metals

Understanding electron removal in transition metals like nickel is important when determining the electronic configuration of ions. In transition metals, the removal of electrons follows a distinct pattern due to the close energy levels of the 3d and 4s orbitals.

When an atom of a transition metal forms a positive ion, electrons are typically removed from the 4s orbital first, even though this orbital is filled before the 3d during neutral atom configuration. This is because the 4s orbital becomes higher in energy compared to the 3d after electrons fill the orbitals, making it more energy-efficient to remove from 4s first.

In the case of \(\text{Ni}^{2+}\), two electrons are removed from the 4s orbital, modifying the configuration from\[1s^2 \ 2s^2 \ 2p^6 \ 3s^2 \ 3p^6 \ 4s^2 \ 3d^8\]to\[1s^2 \ 2s^2 \ 2p^6 \ 3s^2 \ 3p^6 \ 3d^8\].This electron removal prioritization is unique to transition metals and essential for accurately determining the properties of metal cations.