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" The sum of "ns" and "(n-1)d" electrons...

" The sum of "ns" and "(n-1)d" electrons in Tc are : "

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The transition element ( with few exceptions ) show a large number of oxidation states . The various oxidation states are related to the electronic configuration of their atoms. The variable oxidation states of a transition metal is due to the involvement of (n-1)d and outer ns electrons . For the first five elements of 3d transition series , the minimum oxidation state is equal to the number of electrons in 4s shell and the maximum oxidation state is equal to the sum of 4s and 3d electrons. The relative stability of various oxidation states of a given element can be explained on the basis of stability of d^(0),d^(5) and d^(10) configuration . In 3d series, the maximum oxidation state is shown by

Assertion (A): Transition metals have low melting points. Reason (R): The involvement of greater number of (n-1)d and ns electrons in the interatomic metallic bonding.

Many transition elements also form 3+ ions by losing one (n-1)d electron in addition to the two ns electrons except

Assertion: Mn atom loses ns electrons first during ionisation as compared to (n-1) d electrons Reason: The effective nuclear charge experienced by (n-1) d electrons is greater than that by ns electrons.

Assertion : Ionisation of transition metals involves loss of ns electron before (n - 1)d electrons. Reason : Filling of ns-orbitals takes placed before (n - 1)d orbitals.