Question:

$\left[ Ni ( CO )_{4}\right]$ is a diamagnetic complex. All the electrons in the complex are paired.

Updated On: Jul 28, 2022
  • Both Assertion and Reason are true and Reason is the correct explanation of Assertion
  • Both Assertion and Reason are true and Reason is not the correct explanation of Assertion
  • Assertion is true but Reason is false
  • Both Assertion and Reason are false
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The Correct Option is A

Solution and Explanation

In nickel carbonyl, nickel is in zero oxidation state and contains no unpaired electrons.
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Concepts Used:

Properties of D Block Elements

  • Multiple oxidation states- The oxidation states of d block elements show very few energy gaps; therefore, they exhibit many oxidation states. Also, the energy difference between s and d orbital is very less. Therefore both the electrons are involved in ionic and covalent bond formation, which ultimately leads to multiple oxidation states.
  • Formation of complex compounds- Ligands show a binding behavior and can form so many stable complexes with the help of transition metals. This property is mainly due to:
    • Availability of vacant d orbitals.
    • Comparatively small sizes of metals.
  • Hardness- Transition elements are tough and have high densities because of the presence of unpaired electrons.
  • Melting and boiling points- Melting and boiling points of transition are very high because of the presence of unpaired electrons and partially filled d orbitals. They form strong bonds and have high melting and boiling points.
  • Atomic radii- The atomic and ionic radius of the transition elements decreases as we move from Group 3 to group 6. However, it remains the same between group 7 and group 10, and from group 11 to group 12 increases.
  • Ionization enthalpy- The ionization enthalpies of the transition elements are generally on the greater side as compared to the S block elements