Addressing in hex file

a byte is a byte and has 8 bit

if you address bytewide then you get access to 8 bit per access only

if you want to access a word (16bit) then your address must be aglined to 16bit boundaries

so possible adr is 0x0 0x1 0x2 (from 16bit point of view) what is 0x0 0x2 0x4 in bytewide view

same in 32bit. the lsb of the 32bit data is the lsb of 8bit data

but the msb of your 32bit is the msb of the 4. data byte

think of an array and let the contect be {0,1,2,3,4,5,6,7,8,9,....}

accessing this memory bytewide you will see at

adr[0] 0x0

adr[1] 0x1

adr[2] 0x2

adr[3] 0x3

adr[4] 0x4

but now in 16 bit access mode

adr[0] 0x0100

adr[2] 0x0302

adr[4] 0x0504

and in 32 bit

adr[0] 0x03020100

adr[4] 0x07060504

if you write down the possible addresses in binary maybe this helps you

8 bit 00000 00001 00010 00011 00100 00101

16 bit 00000 00010 00100 00110 01000

32 bit 00000 00100 01000 01100 10000

showing only the lowest 5 adr bits