Question

Which of the following statement(s) are correct?
1. Electronic configuration of $\text{ Cr }$is $\left[ \text{Ar} \right]\text{ 3}{{\text{d}}^{\text{4}}}\text{4}{{\text{s}}^{\text{2}}}\text{ }$(Atomic number of$\text{ Cr = 24 }$).
2. The magnetic quantum number may have a negative value.
3. In the silver atom, 23 electrons have a spin of one type and 24 of the opposite type (Atomic number of$\text{ Ag = 47 }$ ).
4. The oxidation state of nitrogen in $\text{ H}{{\text{N}}_{\text{3}}}\text{ }$ is$\text{ }-3\text{ }$ .
A) 1, 2, 3
B) 2, 3, 4
C) 3, 4
D) 1, 2, 4

Answer 1

it is known that half-filled and the full filled orbitals have greater stability .thus if possible elements attain the exceptional electronic configuration. The magnetic quantum number depends on the azimuthal quantum number. It is written as $\text{ m}l\text{ = }\pm l\text{ }$ , where ‘l’ is the azimuthal quantum number. Each orbital accommodates two electrons. They have opposite spins.

Complete step by step answer:
1. Transition element or the d-block element is the element in which the valence shell electrons enter into the d-orbitals. Chromium is a d-block element. It has an atomic number of 24. the electronic configuration of chromium is written as follows,
$\text{ Cr = }\left[ \text{Ar} \right]\text{ 3}{{\text{d}}^{\text{4}}}\text{4}{{\text{s}}^{\text{2}}}\text{ }$
Thus d-block element attains extra stability when d orbitals are half-filled $\text{ }{{\text{d}}^{\text{5}}}\text{ }$ or filled $\text{ }{{\text{d}}^{\text{10}}}\text{ }$. Thus one electron from the $\text{ 4s }$ orbitals shifts to the $\text{3d }$ orbitals. Thus the exceptional electronic configuration of chromium is as,
$\text{ Cr = }\left[ \text{Ar} \right]\text{ 3}{{\text{d}}^{5}}\text{4}{{\text{s}}^{1}}\text{ }$
2. Magnetic quantum number represents the total number of orbitals and their orientation in space. This is represented by the term $\text{ }{{\text{m}}_{l}}\text{ }$ .the value of magnetic quantum number depend on the azimuthal or the orbital quantum number .the value of $\text{ }{{\text{m}}_{l}}\text{ }$ ranges from the $\text{ }-l\text{ }$ to $\text{ +}l\text{ }$. Therefore magnetic quantum numbers may have a negative value.
3. We know that each orbital accommodates two electrons. This electron has the opposite spin. If one electron has an upward spin and the other electrons have a downward spin.
The atomic number of silver is 47.thus there are a total of 47 electrons in the silver atom. If each orbital accommodate 2 electrons then 46 electrons is accommodated in 23 orbitals $\text{ 23 }\times \text{2 = 46 }{{\text{e}}^{-}}\text{ }$ .this 23 orbitals accommodate 23 electrons of upward spin and 23 electrons with downward spin. There is one more electron left in the silver atom which is placed in 24 orbits.
4. Oxidation state of an atom is defined as the charge on the atom of an element. Oxidation number refers to the number of electron gain or loss by a species in neutral state, the sum of the oxidation state of atoms in the molecules is equal to zero. Hydrogen is an electropositive element. It has an oxidation state as $\text{ +1 }$ .lets calculate the oxidation state of nitrogen.
$\begin{aligned} & \text{ O}\text{.S}\text{. of }{{\text{N}}_{\text{3}}}\text{H = 3}\times \text{O}\text{.S}\text{. of N +}\left( \text{O}\text{.S}\text{.of H} \right) \\\ & \Rightarrow \text{ 0 = 3}\times \text{O}\text{.S}\text{. of N + }1 \\\ & \therefore \text{O}\text{.S}\text{. of N = }-\dfrac{1}{3}\text{ } \\\ \end{aligned}$
Thus, statements 1, 2, and 3 are correct.

Hence, (A) is the correct option.

Note: note that, if each orbital accommodate two electrons then we can establish a relation between the number of orbitals and the number of electrons as follows,
For an even number of electrons,
$\text{ Number of orbitals = }\dfrac{\text{Total number of electrons}}{\text{2}}\text{ }$
For an odd number of electrons add one to the relation is, $\text{ Number of orbitals = }\dfrac{\text{Total number of electrons}}{\text{2}}\text{ + 1 }$