To solve the problem, we need to identify the relationship between the complex ions \([Co(NH_3)_5(NO_2)]^{2+}\) and \([Co(NH_3)_5(ONO)]^{2+}\).
1. Analyzing the Complex Ions:
Both complexes have the same formula: Co, 5 NH\(_3\) ligands, and a nitro group, with a 2+ charge. The difference lies in the ligand:
- In \([Co(NH_3)_5(NO_2)]^{2+}\), the NO\(_2\) ligand is coordinated via the nitrogen atom (nitro, -NO\(_2\)).
- In \([Co(NH_3)_5(ONO)]^{2+}\), the NO\(_2\) ligand is coordinated via an oxygen atom (nitrito, -ONO).
2. Identifying the Type of Isomerism:
The two complexes have the same molecular formula but differ in how the NO\(_2\) ligand is attached to the metal center. This type of isomerism, where the same ligand binds through different atoms, is called linkage isomerism. The NO\(_2\)/ONO\(^-\) ligand is ambidentate, meaning it can coordinate via either the N or O atom, leading to these two forms.
3. Considering Other Types of Isomerism:
- Coordination isomerism involves swapping ligands between cation and anion, but here both are cations with the same ligands.
- Geometric isomerism (e.g., cis-trans) is not applicable, as \([Co(NH_3)_5X]\) (with 5 identical ligands) lacks the necessary symmetry for such isomerism.
Linkage isomerism is the correct classification.
Final Answer:
The complex ions \([Co(NH_3)_5(NO_2)]^{2+}\) and \([Co(NH_3)_5(ONO)]^{2+}\) are called linkage isomers.
A racing track is built around an elliptical ground whose equation is given by \[ 9x^2 + 16y^2 = 144 \] The width of the track is \(3\) m as shown. Based on the given information answer the following: 
(i) Express \(y\) as a function of \(x\) from the given equation of ellipse.
(ii) Integrate the function obtained in (i) with respect to \(x\).
(iii)(a) Find the area of the region enclosed within the elliptical ground excluding the track using integration.
OR
(iii)(b) Write the coordinates of the points \(P\) and \(Q\) where the outer edge of the track cuts \(x\)-axis and \(y\)-axis in first quadrant and find the area of triangle formed by points \(P,O,Q\).