Total number of unpaired electron(s) present in both cationic and anionic part of compound . β Chemical Bonding Chemistry Question
Question
Total number of unpaired electron(s) present in both cationic and anionic part of compound $O_2[PtF_6]$.

Answer: 2<BR><BR>**SOLUTION:** $O_2^+[PTF_6]^-$<BR>$O_2^+$ HAS ONE UNPAIRED $E^-$.<BR>$[PTF_6]^-$ HAS ONE UNPAIRED $E^-$, BECAUSE PT IS IN +5 OXIDATION STATE SO TOTAL UNPAIRED ELECTRON IS 2.
π‘ Solution & Explanation
Step 1: The dimer of BH3 is diborane (B2H6). In diborane, each boron atom forms four bonds (two terminal B-H 2c-2e bonds and two bridging B-H-B 3c-2e bonds) using four sp3 hybrid orbitals. Step 2: The dimer of BeH2 is Be2H4. In Be2H4, each beryllium atom is connected to two terminal hydrogens and two bridging hydrogens (forming a planar ring) using sp2 hybrid orbitals. Step 3: Thus, the hybridization of the central atoms in the dimers of BH3 and BeH2 is sp3 and sp2 respectively, matching option (b).
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