NEET Practice Questions (MCQs) with Answers & Solutions

German silver is an alloy composed of Copper (Cu), Zinc (Zn), and Nickel (Ni). It is known for its silver-white appearance and is used in various applications such as cutlery and musical instruments.

To find the oxidation state of Cr in $K_2Cr_2O_7$, we can set up an equation based on the known oxidation states of K and O. Potassium (K) has an oxidation state of +1, and oxygen (O) has an oxidation state of -2. Let the oxidation state of Cr be x. The sum of the oxidation states in a neutral compound is zero.

For $K_2Cr_2O_7$:

$2(+1) + 2(x) + 7(-2) = 0$

$2 + 2x - 14 = 0$

$2x - 12 = 0$

$2x = 12$

$x = +6$

Therefore, the oxidation state of Cr in $K_2Cr_2O_7$ is +6.

Europium (Eu) is a lanthanide element that commonly exhibits +2 and +3 oxidation states. This is due to its electronic configuration, which allows it to achieve stability in both oxidation states.

All the given compounds $TiCl_3$, $FeCl_3$, and $CoCl_2$ are colored. This is because these compounds contain transition metals with unpaired electrons in their d-orbitals, which can absorb visible light and cause the compounds to appear colored.

Transition elements are known to exhibit variable valencies due to the involvement of (n-1)d and ns electrons in bonding. This is a characteristic property of transition metals.

The element with atomic number 56 is Barium (Ba), which belongs to the group of alkaline earth metals in the periodic table.

German silver is an alloy composed primarily of copper, zinc, and nickel. It does not contain silver, hence the percentage proportion of silver in german silver is 0%.

Both Lanthanoids and Actinides exhibit variable oxidation states. This is due to the fact that their electrons can occupy f-orbitals, which allows them to exhibit multiple oxidation states. This property is more pronounced in Actinides compared to Lanthanoids.

The magnetic moment ( ext{in BM}) is calculated using the formula $ ext{μ}_{ ext{eff}} = ext{√}(n(n+2))$ where n is the number of unpaired electrons. For $Mn^{2+}$ (d^5), $Cr^{2+}$ (d^4), and $V^{2+}$ (d^3), the number of unpaired electrons are 5, 4, and 3 respectively. Thus, the order is $Mn^{2+} > Cr^{2+} > V^{2+}$.

The stability of $Gd^{3+}$ ion is due to its half-filled 4f orbital, which provides extra stability according to Hund's rule. A half-filled 4f orbital configuration means that it has exactly 7 electrons, which is a stable arrangement due to maximum spin multiplicity.