​in Q.1 answer is 4 but how is it possible since all other than (BO3^3- ) do not have electron to enite ? explain
​in Q.1 answer is 4 but how is it possible since all other than (BO3^3- ) do not have electron to enite ? explain • +4 If only t u e correspondi • Mv•ks O In all other cases. O 9 3i-$rq ä, 45TFTT I Number of radicals which@roduce coloyr on flame Na• , , A13%Bo-g3-vBa2+1 Li+ •O , B03- Ba2+ Li+ 2. Number of halide ions which change their oxidation number on heating with Mn02 + conc.H2SO Mn02 + H2S04 ow many monochlorinated products formed when S-2 chlorobutane undergo reacti •th Cid'hv s-2 C12 / t I Space for Rough Work /

Dear Student,

Flame is produced due to electron transitions from excited state to normal state. Electrons get excited due to heat energy from the flame. When they go back to the ground state, they release electromagnetic energy which may have a characteristic colour.
Inner shell electrons can take part in the transitions too.

From the given options, all except Al3+ and Mg2+ show characteristic colours under the flame test.

It is generally shown by s-block elements, though magnesium s an exception.

Thus, the correct answer is four.

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