The number of correct options is (a) I ⊖>Br⊖>Cl⊖>F⊖ (polarisability)  (b) Li⊕>Na⊕>K⊕>Rb⊕ (polarisation power)  (c) H2O>H2S>H2Se>H2 Te (order of b.pt.)  (d) H2⊖

(a), (b) and (d) are correct options.(a) I⊖>Br⊖>Cl⊖>F (Polarisability).According to Fajan's rule, large anion is more polarisable.(b) Li⊕>Na⊕>K⊕>Rb⊕ (Polarisation power).According to Fajan's rule, small cation is more polarisable.(c) Wrong.      Boiling point order is(d) Both H2⊖ and H2⊕ have same bond order of 0.5.     H2⊕ is more stable than H2⊖, since H2⊕contains     one e- in bonding orbitals whereas H2⊖ contains     one e- in antibonding orbitals.