The correct order of 2nd ionisation energy ?Mn > Cu > Ti > VLi > Na < KCr > Mn > V > TiSi > Mg > Na > Al

The order of second ionization energy generally increases across a period (from left to right) on the periodic table and decreases down a group (from top to bottom). This is because ionization energy is the energy required to remove an electron from an atom. As you move across a period, the number of protons in the nucleus increases, which means the positively charged nucleus attracts the negatively charged electrons more strongly, thus requiring more energy to remove an electron. As you move down a group, the outermost electrons are further from the nucleus and are therefore less strongly attracted to it, thus requiring less energy to remove.

However, there are exceptions to this trend due to electron configuration. For example, elements with a full or half-full subshell have higher ionization energies because these configurations are more stable.

Given the elements you've listed, the correct order of second ionization energy from highest to lowest should be:

Cu > Mn > Cr > V > Ti > Si > Al > Mg > Na > K > Li

Here's why:

• Copper (Cu) has the highest second ionization energy because it has a full 3d10 subshell after the first ionization.
• Manganese (Mn), Chromium (Cr), Vanadium (V), and Titanium (Ti) follow next. They have relatively high second ionization energies because they have half-full or nearly half-full 3d subshells after the first ionization.
• Silicon (Si), Aluminum (Al), and Magnesium (Mg) have lower second ionization energies because they are in the second period of the periodic table, so their outermost electrons are closer to the nucleus and less strongly attracted to it.
• Sodium (Na), Potassium (K), and Lithium (Li) have the lowest second ionization energies because they are in the first group of the periodic table, so their outermost electrons are in a new energy level further from the nucleus and less strongly attracted to it.