Chemistry MCQs for NEET — Practice Questions with Answers

The compound $[Co(H_2O)_6]Cl_2$ dissociates in water to form one $[Co(H_2O)_6]^{2+}$ ion and two $Cl^-$ ions. Thus, the total number of ions produced is 1 + 2 = 3. Therefore, the correct option is 3 ions. Hence, option o2 is correct.

The correct formula for dichlorido bis (urea) copper (II) is $ [CuCl_2(O=C(NH_2)_2)_2] $. In this complex, the copper ion (Cu^2+) is coordinated to two chloride ions (Cl^-) and two urea molecules. This is indicated by the bis (urea) and dichlorido in the name.

The correct formula for diammine silver (I) chloride is $ [Ag(NH_3)_2]Cl $. In this complex, the silver ion (Ag^+) is coordinated to two ammonia molecules (NH_3), forming the diammine complex, and there is one chloride ion (Cl^-) balancing the charge.

The correct formula for sodium nitroprusside is $ Na_2[Fe(CN)_5NO] $. In this complex, the iron (Fe) is in the +3 oxidation state and is coordinated to five cyanide ions (CN^-) and one nitrosyl group (NO). There are two sodium ions (Na^+) to balance the charge of the complex.

To determine the oxidation state of chromium in $Na_2[CrF_4O]$, we need to consider the charges of the other components in the complex. Sodium (Na) has an oxidation state of +1. Fluoride (F) has an oxidation state of -1, and oxygen (O) usually has an oxidation state of -2. Let the oxidation state of chromium be x. The net charge of the complex ion $[CrF_4O]^{2-}$ should be equal to -2. Therefore, the equation can be set up as follows:

$$2(+1) + x + 4(-1) + (-2) = 0$$

Solving for x, we get:

$$2 + x - 4 - 2 = 0$$

$$x = +4$$

Hence, the oxidation state of chromium in $Na_2[CrF_4O]$ is IV.

The correct IUPAC name of $Fe_4[Fe(CN)_6]_3$ can be determined by breaking down the complex. In this complex, $Fe$ is in two different oxidation states. The $[Fe(CN)_6]^{3-}$ ion is called hexacyanoferrate(III), indicating that the iron in this part of the complex is in the +3 oxidation state. The four $Fe$ atoms outside the brackets balance the charge of the hexacyanoferrate ions. Therefore, the IUPAC name for this complex is Iron(III) hexacyanoferrate(II).

To determine the oxidation state of the central metal atom in various complexes, we need to consider the charges of the ligands and the overall charge of the complex. In $[Cr(CO)_6]$, carbon monoxide (CO) is a neutral ligand, meaning it has an oxidation state of 0. Therefore, the oxidation state of chromium in $[Cr(CO)_6]$ is 0. This makes $[Cr(CO)_6]$ the correct answer where the oxidation state of the metal is zero.

To find the oxidation state of nickel in the complex compound $K_4[Ni(CN)_4]$, we can use the following steps:

1. Let the oxidation state of nickel (Ni) be x.
2. Potassium (K) has an oxidation state of +1, and there are 4 potassium ions, so the total oxidation state contribution from potassium is +4.
3. Cyanide (CN) is a ligand with an oxidation state of -1, and there are 4 cyanide ions, so the total oxidation state contribution from cyanide is -4.
4. The overall charge of the complex ion $[Ni(CN)_4]$ is -4 (since it needs to balance the +4 charge from potassium to make the compound neutral).

So, we have the equation:

$$4(+1) + x + 4(-1) = 0$$

Simplifying this, we get:

$$4 + x - 4 = 0$$

$$x = 0$$

Therefore, the oxidation state of nickel in $K_4[Ni(CN)_4]$ is 0.

The highest possible oxidation states for transition metals are generally found by considering their position in the periodic table. Chromium (Cr) can reach an oxidation state of +6, and manganese (Mn) can reach an oxidation state of +7.

In the compound $CrO_2Cl_2$, chromium is in the +6 oxidation state because oxygen has a -2 oxidation state and chlorine has a -1 oxidation state. For the overall charge to be neutral, Cr must be +6.

In the compound $MnO_4^-$, manganese is in the +7 oxidation state because oxygen has a -2 oxidation state and there are four oxygen atoms contributing a total of -8. For the overall charge to be -1, Mn must be +7.

Thus, both $CrO_2Cl_2$ and $MnO_4^-$ have metals in their highest possible oxidation states.

To determine the number of unpaired electrons in the complex ion $[CoF_6]^{3-}$, we follow these steps:

1. The atomic number of cobalt (Co) is 27, so its electron configuration is $[Ar] 3d^7 4s^2$.
2. In the $[CoF_6]^{3-}$ complex, cobalt is in the +3 oxidation state. This means cobalt loses three electrons, resulting in the electron configuration $[Ar] 3d^6$.
3. Fluoride (F-) is a weak field ligand and does not cause pairing of electrons.
4. In a weak field, the $3d$ orbitals remain unpaired as much as possible.

Thus, the electron configuration for $Co^{3+}$ in a weak field is $t_{2g}^4 e_g^2$, where there are 4 electrons in the $t_{2g}$ orbitals and 2 electrons in the $e_g$ orbitals.

Among the 6 electrons in the $d$ orbitals, 4 will remain unpaired.

Therefore, the number of unpaired electrons in $[CoF_6]^{3-}$ is 4.