According to the Wikipedia link of IBM hexadecimal floating point:
Consider encoding the value −118.625 as an IBM single-precision floating-point value.
The value is negative, so the sign bit is 1.
The value 118.62510 in binary is 1110110.1012. This value is normalized by moving the radix point left four bits (one hexadecimal digit) at a time until the leftmost digit is zero, yielding 0.011101101012. The remaining rightmost digits are padded with zeros, yielding a 24-bit fraction of .0111 0110 1010 0000 0000 00002.
The normalized value moved the radix point two digits to the left, yielding a multiplier and exponent of 16+2. A bias of +64 is added to the exponent (+2), yielding +66, which is 100 00102.
Combining the sign, exponent plus bias, and normalized fraction produces this encoding:
S Exp Fraction
1 100 0010 0111 0110 1010 0000 0000 0000
In other words, the number represented is −0.76A00016 × 1666 − 64 = −0.4633789… × 16+2 = −118.625
Now, the definition of normalization according to Wikipedia says that
In base $b$ a normalized number will have the form $±d_0.d_1d_2d_3...×b_n$ where $d_0≠0$, and the digits $d_0,d_1,d_2,d_3,...$ are integers between $0$ and $b−1$
So, how is $0.011101101012 \times 16^2$ a normalized number?
In fact this number cannot be represented as a normalized one with base $16$ exponent because the closest we can get is $1.1101101012 \times 16^1 \times 2^2$. What am I missing here?