The wave function, ψn, l, ml is a mathematical function whose value depends upon spherical polar coordinates r, θ, ϕ of the electron and characterized by the quantum numbers n, l and ml . Here r is the distance from the nucleus, θ is colatitude and ϕ is azimuth. In the mathematical functions given in the Table, Z is the atomic number and ao is Bohr radius Column 1Column 2Column 3(I) 1s orbital\begin{array}{l} ext { (i) } \psi_{ n , 1, m_1} \ \propto\left(\frac{ Z }{ a _{ o }} ight)^{\frac{3}{2}} e ^{-\left(\frac{ Zr }{ e _{ o }} ight)}\end{array}(P) (II) 2s orbital(ii) One radial node(Q) Probability density at the nucleus \propto \frac{1}{ a _{ o }^3}.(III) 2 p _{ z } orbital\begin{array}{l} ext { (iii) } \psi_{ n }, 1, m_{ l } \ \propto\left(\frac{ Z }{ a _{ o }} ight)^{\frac{5}{2}} re ^{-\left(\frac{ Zr }{2 a _{ o }} ight)} \cos heta\end{array}(R) Probability density is maximum at the nucleus.(IV) 3 d_{ z }^2 orbital(iv) x y-plane is a nodal plane(S) Energy needed to excite an electron from n =2 state to n =4 state is \frac{27}{32} times the energy needed to excite are electron from n =2 state to n =6 state.For the given orbital in column 1, the only correct combination for any hydrogen-like species is

Question

The wave function, ψn, l, ml is a mathematical function whose value depends upon spherical polar coordinates r, θ, ϕ of the electron and characterized by the quantum numbers n, l and ml . Here r is the distance from the nucleus, θ is colatitude and ϕ is azimuth. In the mathematical functions given in the Table, Z is the atomic number and ao is Bohr radius Column 1Column 2Column 3(I) 1s orbital\begin{array}{l}	ext { (i) } \psi_{ n , 1, m_1} \ \propto\left(\frac{ Z }{ a _{ o }}ight)^{\frac{3}{2}} e ^{-\left(\frac{ Zr }{ e _{ o }}ight)}\end{array}(P) (II) 2s orbital(ii) One radial node(Q) Probability density at the nucleus \propto \frac{1}{ a _{ o }^3}.(III) 2 p _{ z } orbital\begin{array}{l}	ext { (iii) } \psi_{ n }, 1, m_{ l } \ \propto\left(\frac{ Z }{ a _{ o }}ight)^{\frac{5}{2}} re ^{-\left(\frac{ Zr }{2 a _{ o }}ight)} \cos 	heta\end{array}(R) Probability density is maximum at the nucleus.(IV) 3 d_{ z }^2 orbital(iv) x y-plane is a nodal plane(S) Energy needed to excite an electron from n =2 state to n =4 state is \frac{27}{32} times the energy needed to excite are electron from n =2 state to n =6 state.For the given orbital in column 1, the only correct combination for any hydrogen-like species is

Accepted Answer

(D) (II)&#160;(ii)&#160;(P)

Answer

(A) (III)&#160;(iii)&#160;(P)

Answer

(B) (I)&#160;(ii)&#160;(S)

Answer

(C) (IV)&#160;(iv)&#160;(R)

Answer & Solution

Step-by-step Solution

See the Complete Solution

Column 1	Column 2	Column 3
(I) 1s orbital	\(\begin{array}{l}\text { (i) } \psi_{ n , 1, m_1} \\ \propto\left(\frac{ Z }{ a _{ o }}\right)^{\frac{3}{2}} e ^{-\left(\frac{ Zr }{ e _{ o }}\right)}\end{array}\)	(P)
(II) 2s orbital	(ii) One radial node	(Q) Probability density at the nucleus \(\propto \frac{1}{ a _{ o }^3}\).
(III) \(2 p _{ z }\) orbital	\(\begin{array}{l}\text { (iii) } \psi_{ n }, 1, m_{ l } \\ \propto\left(\frac{ Z }{ a _{ o }}\right)^{\frac{5}{2}} re ^{-\left(\frac{ Zr }{2 a _{ o }}\right)} \cos \theta\end{array}\)	(R) Probability density is maximum at the nucleus.
(IV) \(3 d_{ z }^2\) orbital	(iv) \(x y\)-plane is a nodal plane	(S) Energy needed to excite an electron from \(n =2\) state to \(n =4\) state is \(\frac{27}{32}\) times the energy needed to excite are electron from \(n =2\) state to \(n =6\) state.