Can you cover this grid?

If that tiling is possible, one needs 9 tetrominoes to do so. Imagine that each column is alternately colored black and white. Notice that each tetromino will tile either 3 black tiles and 1 white tile OR 3 white tiles and 1 black tile. We know that there are 18 black and 18 white tiles, so we will let $k$ be the number of tetrominoes covering 3 black and 1 white tile while $(9-k)$ covering 1 black and 3 white tiles. We can set up a system of equations that governs how the tiles are covered like so: $3k+(9-k)=18 \text{ black tiles }$ and $k+(9-k)(3)=18 \text{ white tiles }.$ Both of these equations are unsolvable in integers, so it is not possible to tile the board in this way.

We will prove it is impossible.

Assume the columns are alternately coloured black and white. There are 18 black and 18 white squares.

Since there are 36 squares in the 6x6 and a tetromino has 4 squares, we need 9 tetrominoes.

Each tetromino will cover either 3 white squares or 1 white square no matter how you place it, and all 18 white squares must be covered. We have 9 tetrominoes. It is impossible for 9 odd numbers to add up to 18 which is even. Hence there is no way to place the tetrominoes.

I solve this problem joining pairs of tetrominoes creating single groups of rectangles of 8 squares of area. Putting them in the big square of 10x10 we will need 12 rectangles(24 tetrominoes), however we will have left an square of 2x2 that we can't fill with a single tetrominoe