The oxidation states displayed by the first half of the first row of transition metals are given in the table below
| Oxidation state | Sc | Ti | V | Cr | Mn |
| +2 | +2 | +2 | +2 | ||
| +3 | +3 | +3 | +3 | +3 | |
| +4 | +4 | +4 | +4 | ||
| +5 | +5 | +6 | |||
| +6 | +7 |
It can be easily observed that except Sc, all others metals display +2 oxidation state. Also, on moving from Sc to Mn, the atomic number increases from 21 to 25. This means the number of electrons in the 3d-orbital also increases from 1 to 5.
Sc (+2) = d1
Ti (+2) = d2
V (+2) = d3
Cr (+2) = d4
Mn (+2) = d5
+2 oxidation state is attained by the loss of the two 4selectrons by these metals. Since the number of delectrons in
(+2) state also increases from Ti(+2) to Mn(+ 2), the stability of +2 state increases (as d-orbital is becoming more and
more half-filled). Mn (+2) has d 5electrons (that is half-filled dshell, which is highly stable).
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 \(P,O,Q\).