Consider the following changes : M(s) - M(g) ...(1) ; M(s) - M2+(g) + 2e- ...(2) ; M(g) - M+(g) + e- ...(3) ; M+(g) - M2+(g) + e- ...(4) ; M(g) - M2+(g) + 2e- ...(5). The second ionization energy

Question

Consider the following changes : M(s) -> M(g) ...(1) ; M(s) -> M2+(g) + 2e- ...(2) ; M(g) -> M+(g) + e- ...(3) ; M+(g) -> M2+(g) + e- ...(4) ; M(g) -> M2+(g) + 2e- ...(5). The second ionization energy of M could be calculated from the energy values associated with :

Prashant Kotian · Accepted Answer

Correct answer: D. Step 1: Identify the element with the given configuration. The atomic number is 36 (Kr) + 10 (4d) + 14 (4f) + 2 (5s) + 6 (5p) + 2 (6s) = 70, which is Ytterbium (Yb). Step 2: Ytterbium is a lanthanide element. The last electron entered and completed the 4f subshell (4f14). Step 3: Elements in which the f-orbitals are being filled belong to the f-block, so the correct option is (d).

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