AP EAMCET · Chemistry · Classification of Elements and Periodicity in Properties

The ground state energy of $\mathrm{H}$ is equal to

A First excited state energy of $\mathrm{He}^{+}$
B Ground state energy of $\mathrm{Be}^{3+}$
C First excited state energy of $\mathrm{Li}^{2+}$
D Ground state energy of $\mathrm{Li}^{2+}$

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Answer & Solution

Correct Answer

(A) First excited state energy of $\mathrm{He}^{+}$

Step-by-step Solution

Detailed explanation

No solution. Refer to answer key.

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Match the following elements, given as hints to their first, second ionization enthalpies and
electron gain enthalpies.

List- $I$	List- $II$
$i)$ Most reactive non-metal	$A)$ $∆ H_{1} = 419, ∆ H_{2} = 3051, ∆_{Eg} H = - 48$
$ii)$ Most reactive metal	$B)$ $∆ H_{1} = 1681, ∆ H_{2} = 3374, ∆_{Eg} H = - 328$
$iii)$ Least reactive element	$C)$ $∆ H_{1} = 738, ∆ H_{2} = 1451, ∆_{Eg} H = - 40$
$iv)$ Metal forming binary halide	$D)$ $∆ H_{1} = 3272, ∆ H_{2} = 5251, ∆_{Eg} H = + 48$

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The ground state energy of \(\mathrm{H}\) is equal to

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