Identify Incorrect Orbital Combination for Helium Ion

To determine the INCORRECT combination for the $\text{He}^+$ ion, let us analyze the relationships between the entries in Column 1, Column 2, and Column 3 for each given option.

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1. Analysis of the Wave Functions and Properties

• 1s Orbital $\text{(I)}$:

  • The wave function for a $1s$ orbital ($n=1, l=0, m_l=0$) of a hydrogen-like species ($\text{He}^+$, $Z=2$) is given by: $\psi_{1s} = \frac{1}{\sqrt{\pi}} \left(\frac{Z}{a_o}\right)^{\frac{3}{2}} e^{-\left(\frac{Zr}{a_o}\right)}$ This directly matches (i) in Column 2: $\psi_{n,l,m_l} \propto \left(\frac{Z}{a_o}\right)^{\frac{3}{2}} e^{-\left(\frac{Zr}{a_o}\right)}$

  • The probability density $|\psi_{1s}|^2 \propto e^{-\left(\frac{2Zr}{a_o}\right)}$ reaches its absolute maximum at $r = 0$ (at the nucleus). This matches (R) in Column 3.

• $2p_z$ Orbital $\text{(III)}$:

  • The wave function for a $2p_z$ orbital ($n=2, l=1, m_l=0$) is given by: $\psi_{2p_z} \propto \left(\frac{Z}{a_o}\right)^{\frac{5}{2}} r e^{-\left(\frac{Zr}{2a_o}\right)} \cos\theta$ This corresponds to (iii) in Column 2.

• 2s Orbital $\text{(II)}$:

  • Number of radial nodes = $n - l - 1 = 2 - 0 - 1 = 1$. This corresponds to (ii) in Column 2.
  • For any s-orbital ($l=0$), the probability density at the nucleus ($r=0$) is non-zero and proportional to $\left(\frac{Z}{a_o}\right)^3 \propto \frac{1}{a_o^3}$. This corresponds to (Q) in Column 3.

• Excitation Energy Statement $\text{(S)}$:

  • For a single-electron species like $\text{He}^+$, the energy required to excite an electron between energy levels $n_1$ and $n_2$ is given by: $\Delta E = R_y Z^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$
  • For transition $n = 2 \to n = 4$: $\Delta E_{2 \to 4} = R_y Z^2 \left(\frac{1}{2^2} - \frac{1}{4^2}\right) = R_y Z^2 \left(\frac{1}{4} - \frac{1}{16}\right) = \frac{3}{16} R_y Z^2$
  • For transition $n = 2 \to n = 6$: $\Delta E_{2 \to 6} = R_y Z^2 \left(\frac{1}{2^2} - \frac{1}{6^2}\right) = R_y Z^2 \left(\frac{1}{4} - \frac{1}{36}\right) = \frac{8}{36} R_y Z^2 = \frac{2}{9} R_y Z^2$
  • The ratio of energies is: $\frac{\Delta E_{2 \to 4}}{\Delta E_{2 \to 6}} = \frac{\frac{3}{16}}{\frac{2}{9}} = \frac{27}{32}$ Hence, statement (S) is correct for $\text{He}^+$.

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2. Evaluation of the Options

1. Option A: (I) (i) (R)

   • (I) is 1s orbital $\rightarrow$ (i) is its correct wave function $\rightarrow$ (R) correctly states maximum probability density at the nucleus.
   • Result: CORRECT combination.

2. Option B: (II) (ii) (Q)

   • (II) is 2s orbital $\rightarrow$ (ii) correctly gives one radial node $\rightarrow$ (Q) correctly gives probability density at the nucleus $\propto \frac{1}{a_o^3}$.
   • Result: CORRECT combination.

3. Option C: (I) (iii) (R)

   • (I) represents the 1s orbital, whereas (iii) is the wave function expression for the $2p_z$ orbital.
   • A 1s orbital cannot be represented by wave function (iii).
   • Result: INCORRECT combination.

4. Option D: (I) (i) (S)

   • (I) is 1s orbital $\rightarrow$ (i) is its correct wave function $\rightarrow$ (S) is a true energy transition relationship for $\text{He}^+$.
   • Result: CORRECT combination.

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Conclusion

The only INCORRECT combination is (C).