Here we have to get the symbol of the selenium ion having two extra electrons.
The symbol of the selenium ion with two extra electrons is
.
The number of electrons present in selenium is 34. The electronic configuration of selenium (Se) is 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁴.
Now we can see that there are 4 electrons in p-orbital. The p-orbital can take maximum 6 electrons thus there is two vacant space in 4p orbital for selenium. If we add two electrons then the total number of electrons will be 36. The electronic configuration can be written as:
[Se²⁻] = 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶.
The two negative charge on selenium makes it an anion as one electron always posses one unit negative charge. Thus for the incorporation of the two electrons there will be accumulation of two negative charge on the selenium atom and it will be expressed as Se²⁻.
Answer: (1) It is exothermic and DH equals -91.8 kJ.
Explanation: 
There are two types of reactions in terms of heat:
Endothermic reactions are those reactions in which heat is absorbed by the system and exothermic reactions are those reactions in which heat is released by the system.
As the heat is given on the products side, it means the heat is released in the reaction and thus the reaction is exothermic. The enthalpy change for exothermic reaction is written as negative and the enthalpy change for endothermic reaction is written as positive.
The concentration in mol/L of the chemist's potassium iodide solution : = 4.4 x 10⁻⁴ M
<h3>Further explanation</h3>
Given
88. μmol of Potassium iodide
200. mL volumetric flask
Required
The concentration
Solution
Molarity is a way to express the concentration of the solution
Molarity shows the number of moles of solute in every 1 liter of solute or mmol in each ml of solution

Conversion :
88. μmol = 88 . 10⁺⁶ mol = 8.8 10⁻⁵ mol
200 ml = 0.2 L
The molarity :
= n : V
= 8.8.10⁻⁵ : 0.2
= 4.4 x 10⁻⁴ M
Answer:
n=6 to n=3 (B)
Explanation:
Energy of an electron present in the
orbit is directly proportional to
.Hence a transistion from one orbit to another orbit emits an energy proportional to the difference of their squares of the orbits. that is if an electron travels from orbit n1 to orbit n2 then it emits an energy corresponding to
.So in the above question the highest energy emission occurs when an electron moves from n=6 to n=3.(Highest difference of energy levels).
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