Highest oxidation state of manganese in fluoride is +4 (Mn) but the highest oxidation state in oxides is +7 (Mn2O7). The reason is :
1. | Fluorine is more electronegative than oxygen. |
2. | Fluorine does not possess d orbitals. |
3. | Fluorine stabilizes lower oxidation state of Mn. |
4. | In covalent compounds, fluorine can only form a single bond while oxygen forms a double bond. |
The metallic radii of some transition elements are given below. An element that has the highest density is:
Element | Fe | Co | Ni | Cu |
Metallic radii/pm | 126 | 125 | 125 | 128 |
1. Fe
2. Ni
3. Co
4. Cu
Among the following, which are amphoteric oxides?
1. V2O5, Cr2O3
2. Mn2O7, CrO3
3. CrO, V2O5
4. V2O5, V2O4
The electronic configuration of gadolinium (atomic number of 64) is:
1.
2.
3.
4.
Match Column I (Catalyst) with Column II (Process), and mark the appropriate option:
Column I - (Catalyst) | Column II - (Process) | ||
A. | Ni in the presence of hydrogen | 1. | Contact process |
B. | Cu2Cl2 | 2. | Vegetable oil to ghee |
C. | V2O5 | 3. | Sandmeyer reaction |
D. | Finely divided iron | 4. | Haber's process |
5. | Decomposition of KClO3 |
Codes:
A | B | C | D | |
1. | 3 | 4 | 2 | 5 |
2. | 2 | 3 | 1 | 4 |
3. | 5 | 4 | 3 | 2 |
4. | 4 | 5 | 3 | 2 |
Urea reacts with water to form A, which decomposes to form B. When B is passed through (aq), a deep blue-colored solution C is formed
The formula of C is:
2.
3.
4.
Match the following characteristics to the appropriate metal:
Column I - Aspects | Column II - Metal | ||
(a) | The metal which reveals the maximum number of oxidation states | (i) | Scandium |
(b) | The metal although placed in 3d block is considered not as a transition element | (ii) | Copper |
(c) | The metal which does not exhibit variable oxidation states | (iii) | Manganese |
(d) | The metal which in +1 oxidation state in aqueous solution undergoes disproportionation | (iv) | Zinc |
Select the correct option :
(a) | (b) | (c) | (d) | |
1. | (i) | (iv) | (ii) | (iii) |
2. | (iii) | (iv) | (i) | (ii) |
3. | (iii) | (i) | (iv) | (ii) |
4. | (ii) | (iv) | (i) | (iii) |
Actinoids show a greater number of oxidation states than lanthanoids because:
1. | 4f- orbitals are more diffused than 5f-orbitals |
2. | Lesser energy difference between 5f, and 6d orbitals than between 4f, and 5d orbitals |
3. | More energy difference between 5f, and 6d orbitals than between 4f, and 5d orbitals |
4. | More reactive nature of the actinoids than the lanthanoids. |
The lanthanide contraction is responsible for the fact that:
1. Zr and Y have about the same radius .
2. Zr and Zn have the same oxidation state.
3. Zr and Hf have about the same atomic radius .
4. Zr and Nb have a similar oxidation state.
The oxidation state of chromium in the final product formed by the reaction between KI and acidified potassium dichromate solution will be:
1. +2
2. +3
3. +4
4. +6