Question

determine the oxidation state for each of the elements below.
the oxidation state of carbon in oxalic acid ( h_2c_2o_4 ) is.
the oxidation state of bromine in magnesium bromide ( mgbr_2 ) is.
the oxidation state of silicon in silicate ion ( sio_3^{2 - } ) is.

Explanation:

Step1: Oxidation state of carbon in \(H_2C_2O_4\)

Let the oxidation state of \(C\) be \(x\).
Oxidation state of \(H = + 1\) and oxidation state of \(O=-2\).
For \(H_2C_2O_4\), using the formula: \(2\times(+1)+2x + 4\times(-2)=0\) (since the overall charge of the neutral molecule is \(0\)).

$$2 + 2x-8 = 0$$

$$2x=6$$

$$x = + 3$$

Step2: Oxidation state of bromine in \(MgBr_2\)

Let the oxidation state of \(Br\) be \(y\).
Oxidation state of \(Mg = + 2\).
For \(MgBr_2\), using the formula: \(+2+2y = 0\) (since the overall charge of the neutral molecule is \(0\)).

$$2y=-2$$

$$y=-1$$

Step3: Oxidation state of silicon in \(SiO_3^{2 -}\)

Let the oxidation state of \(Si\) be \(z\).
Oxidation state of \(O=-2\).
For \(SiO_3^{2-}\), using the formula: \(z+3\times(-2)=-2\) (since the overall charge of the ion is \(-2\)).

$$z - 6=-2$$

$$z=+4$$

Answer:

The oxidation state of carbon in \(H_2C_2O_4\) is \(+3\).
The oxidation state of bromine in \(MgBr_2\) is \(-1\).
The oxidation state of silicon in \(SiO_3^{2-}\) is \(+4\).