Some Applications Of D And F Block Elements

Since Zn is not a transition element so transition element having lowest atomisation energy out of Cu, V, Fe is Cu.

Step 1: Understanding the Question:
This question asks to determine the acidic/basic nature of two transition metal oxides, V O and CrO.
Step 2: Key Formula or Approach:
The nature of transition metal oxides is generally related to the oxidation state of the metal:
Low oxidation state oxides are typically basic.
Intermediate oxidation state oxides are typically amphoteric.
High oxidation state oxides are typically acidic.
We need to find the oxidation states of Vanadium (V) and Chromium (Cr) in the given oxides.
Step 3: Detailed Explanation:
For V O (Oxide X):
Let the oxidation state of V be 'x'. Oxygen usually has an oxidation state of -2. The oxidation state of Vanadium is +3. For Vanadium, which exhibits states from +2 to +5, +3 is a relatively low state. Oxides in lower oxidation states are basic. For example, V O reacts with acids to form V salts. Therefore, V O is a basic oxide. So, X = basic.
For CrO (Oxide Y):
Let the oxidation state of Cr be 'y'. The oxidation state of Chromium is +2. This is a low oxidation state for Chromium (which commonly exists in +2, +3, and +6 states). Therefore, CrO is a basic oxide. For example, CrO reacts with acids to form Cr salts. So, Y = basic.
Step 4: Final Answer:
Both V O and CrO are basic oxides. Therefore, the correct set is X = basic and Y = basic.

Q59. Atomic number 29 Copper (Cu). Step 1: Electronic configuration Step 2: Number of unpaired electrons Step 3: Spin-only magnetic moment

gets oxidised by into

$C{u}^{2+}+2KI⟶\underset{\text{Unstable}}{Cu{I}_2}\downarrow +2K^{+}$
$I^{-}$is strong R.A it reduces $C{u}^{2+}$ to $C{u}^{+}$
$2Cu{I}_2⟶\underset{\text{(White)’X’}}{C{u}_2{I}_2\downarrow }+I_2$
$KI+I_2⟶K^{+}{I}_3^{-}\text{(Brown solution)}$
$I_3^{-}\rightleftharpoons I_2+I^{-}$
$K{I}_3+N{a}_2{S}_2{O}_3\to KI+\underset{(Y)}{N{a}_2{S}_4{O}_6}$

The correct option is (A): A, B and C only