Q.Why nobelium is the only actinoid with +2 oxidation state?
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Start your 14-day free trial to unlock the full solution →Step 1. Recall nobelium's ground-state configuration. Table 8.14 gives No (Z=102) as [Rn] 5f14 6d0 7s2 -- its 5f subshell is already completely filled in the neutral atom.
Step 2. Form No2+ and compare to the usual actinoid pattern. Removing the two outer 7s electrons gives No2+ = [Rn] 5f14, which keeps the fully-filled f14 subshell completely intact. For a TYPICAL actinoid, by contrast, the dominant, most stable state is +3 (losing a further electron from the f/d subshell on top of the two s electrons, Section 8.14/8.15) -- but for nobelium, removing that third electron would have to come OUT of the already-complete, extra-stable f14 subshell, breaking its special stability.
Step 3. Apply the chapter's own extra-stability principle. The same half-filled/fully-filled subshell stability rule used throughout this chapter (chromium/copper in Section 8.3.1; gadolinium/lutetium in Section 8.12.1) explains why No2+ -- which preserves a full f14 shell -- is unusually favoured over No3+, which would have to break it. This directly mirrors ytterbium among the lanthanoids: Yb2+ is likewise the 'most stable dipositive ion' specifically because it also reaches an f14 configuration (Section 8.12/8.12.2). …
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